Automatic Control Engineering 9789389583748, 9789389872521

Table of contents :
Cover
Half T itle
Title
Copyright
Preface
Table of Content
Chapter 1: Introduction
1.1 The Control System
1.2 Open Loop Control System
1.3 Closed Loop Control System
1.4 Basic Example
1.5 Classifi cation of Control Systems
1.6 General Block Diagram of Closed Loop Control System
1.7 Illustrative Examples of Closed Loop Control System
Chapter 2: Control System Representation
2.1 Introduction
2.2 Transfer Function Representation
2.3 Block Diagram Representation and Reduction
2.4 Signal Flow Graph Representation and Reduction
Chapter 3: Modeling of Control Systems
3.1 Introduction
3.2 Modeling of Mechanical Systems
3.3 Modeling of Electrical Systems
3.4 Thermal Systems
3.5 Hydraulic Systems
Chapter 4: Effects of Feedback on Control Systems
4.1 Introduction
4.2 Effect of Feedback on Output Due to Parameter Variations in Control System
4.3 Effect of Feedback on Sensitivity of Control System
4.4 Effect of Feedback on Gain of Control System
4.5 Effect of Feedback on Dynamics of Control System
4.6 Effect of Feedback on Stability of Control System
4.7 Effect of Feedback on Disturbances in Control Systems
4.8 Conclusions
4.9 Regenerative Feedback
Chapter 5: Time Response Analysis
5.1 Introduction
5.2 Important Terms
5.3 Standard Test Input Signals for Transient Analysis
5.4 Time Response of First Order System
5.5 Time Response of Second Order System
5.6 Time Response Specifi cations of Second Order System
5.7 Design Considerations: Performance Indices
5.8 Concept of Steady State Error
5.9 Effect of Changing Inputs on Steady State Error
5.10 Effect of G(s) H(s) or Type of System on Steady State Error
5.11 Summary of Steady State Errors and Error Constants
Chapter 6: Concepts of Stability and Its Algebraic Solutions
6.1 Introduction
6.2 Concept of Stability
6.3 Types of Stability
6.4 Necessary Condition for Stability
6.5 Hurwitz Stability Criterion
6.6 Routh’s Stability Criterion
6.7 Special Cases of Routh’s Hurwitz Criterion
6.8 Applications of Routh’s Hurwitz Criterion
6.9 Relative Stability Analysis
Chapter 7: Root Locus Technique
7.1 Introduction
7.2 Concept of Root Locus: Map Criterion and Angle Criterion
7.3 Construction Rules of Root Locus
7.4 Determination of K for a Specifi ed Damping Factor d
7.5 Cancellation of Poles and Zeros
7.6 Root Locus When m > n
7.7 Root Contours
Chapter 8: Frequency Response Analysis
8.1 CoRRelation Between Time Response and Frequency Response of a Second Order System
8.2 Frequency Response Specifi cations
8.3 Bode Plot
8.4 Polar Plot
8.5 Nyquist Stability Criterion
8.6 Non-Minimum Phase Systems, All Pass Systems, Minimum Phase Systems
Chapter 9: Control System Components
9.1 Servomotors
9.2 Error Detectors
9.3 Tachogenerators
9.4 Stepper Motors
Chapter 10: Robust Control Systems
10.1 Introduction
10.2 System Sensitivity
10.3 Analysis of Robustness
10.4 PID Controllers
10.5 Tuning of PID Controllers
Chapter 11: Sampled Data Control System
11.1 Introduction
11.2 Components of Sampled Data Control System
11.3 Advantages of Sampled Data Control System Over Analog Control System
11.4 Applications of Sampled Data Control System
11.5 Sampler
11.6 Z-transform
11.7 Hold Circuit/Signal Reconstruction
11.8 Pulse Transfer Function
11.9 Stability Analysis
Appendix I: Laplace Transforms of Some Useful Functions
Answers to Unsolved Exercise (Selected Numericals)
References
Index
Back Cover

Citation preview

Automatic Control Engineering

Aditi Gupta is Assistant Professor in the Department of Electrical & Electronics Engineering at UIET, Panjab University, Chandigarh. She received her B.Tech. degree in Electrical Engineering from MMEC, Mullana, Kurukshetra University, Kurukshetra in 2002 with honours and Masters degree in Electrical Engineering from Panjab University, Chandigarh in 2004 with distinction. She is pursuing Ph.D in Electrical Engineering from Panjab University, Chandigarh. Her research interest includes automatic generation control of power systems, wind power, power system deregulation and congestion management. She has contributed several research papers in national and international journals.

978-93-89583-74-8

1

2

E

c(t) ess r(t)=1

Aditi Gupta • Yajvender Pal Verma

Yajvender Pal Verma received his B.Tech. degree in Electrical Engineering from NIT Hamirpur in 2000 with Honours and Masters degree in Power Systems from Panjab University, Chandigarh in 2002. He is Ph.D in Electrical Engineering from NIT Kurukshetra, Haryana, India. He is presently working as Assistant Professor in the Department of Electrical & Electronics Engineering at UIET, Panjab University, Chandigarh. His research interest includes distributed generation, control and operation of wind power, power system restructuring, congestion management and power system optimisation. He has authored more than 40 research papers in national and international journals of repute and conferences. He has also carried out two research projects sponsored by Department of Science & Technology. He is a member of IEEE, CSI and IAENG.

Automatic Control Engineering

AUTOMATIC CONTROL ENGINEERING

The book presents the topics of control engineering in a relatively easy language, which is of great importance for the beginners. It is compulsory for every student to start with the basic concepts and gradually reach to the competitive level. This book includes competitive objective-type questions at the end of each chapter. Moreover, the major complex topics of control engineering, such as Root Locus Technique, Bode Plot and Nyquist Criterion are explained in an easy manner using step-by-step method. In addition to this, some advanced topics of control engineering, such as sampled data control systems and robust control systems are also included in this book.

TM

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Distributed by: 9 789389 583748 TM

Automatic Control Engineering

Automatic Control Engineering

Aditi Gupta Assistant Professor Department of Electrical and Electronics Engineering UIET Panjab University, Chandigarh

Yajvender Pal Verma Assistant Professor Department of Electrical and Electronics Engineering UIET Panjab University, Chandigarh

©Copyright 2020 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. ISBN: 978-93-89583-74-8 EISBN: 978-93-89872-52-1

Edition: 2020

Preface There have been continuous developments in all areas of engineering applications which have made the role of control system engineering extremely important. This is due to the fact that automation of systems, components and processes is possible only with an effective and efficient control system. Most of the processes which are automated have control system as an integral part. In fact, control systems play an important role in many applications ranging from a small motor speed regulation to robotics control these days. It was this motivation that the authors decided to write a book on automatic control engineering based on their teaching experience. An undergraduate student must be able to understand the basic concepts and applications of conventional and modern control in order to design the control system. Control engineering is taught to most of the engineering disciplines due to its wide application in different spheres of life. This book fulfils the need of an undergraduate and postgraduate student in the course of control and advanced control engineering. This book has been written in simple and understandable language supported with examples and objective type questions for revision besides MATLAB programs. It introduces the concept and analysis of control system along with coverage of latest topic in automatic control like robust control. The reader of this book is expected to have prerequisite knowledge of Laplace transform, differential equations and Z transform. This book contains eleven chapters in all. The basic introduction and concepts are given in chapter 1 with many daily-life examples. Classification of control systems is also presented in this chapter. Mathematical representation of a control system in terms of transfer function, block diagrams and signal flow graphs is presented in chapter 2. It is necessary for further analyzing the control system. Chapter 3 presents the mathematical modeling of electrical systems and other physical systems. A relationship between feedback and other parameters of control system is presented in chapter 4 so as to reveal the importance of feedback in control system. Chapter 5 deals with the time response analysis of control systems which is important for its design and analysis. Comparison of various performance indices for control systems is also given in this chapter. The concepts of stability are progressively built in chapters 6, 7 and 8. Basic concepts of stability like BIBO stability and asymptotic stability are given in chapter 6. Time domain methods of stability analysis like Routh’s-Hurwitz criterion and root locus technique are presented in chapters 6 and 7 respectively. Chapter 8 gives the frequency domain stability analysis of a control system. Many frequency domain stability analysis methods like Bode plot, Polar plot and Nyquist criterion are presented in this chapter in a relative easier and step by step manner. Chapter 9 covers the theory and concepts of various control system components and their examples. Keeping in view the necessity

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Preface

and importance of advanced control systems, chapters 10 and 11 are included in this book. Chapter 10 presents the theory and analysis of robust control system. Chapter 11 deals with the sampled data control systems with many solved examples. The authors would also like to thank the Director, Prof. Renu Vig and faculty colleagues of UIET Panjab University Chandigarh who provided necessary support and encouragement while writing this book. The authors are indebted to their family members who have been a source of inspiration and guidance and also for their patience during the writing and editing of this book. At the end, the authors request all the readers of this book to send their suggestions which will be very valuable in bringing out the improved next edition of this book. Please send your comments and suggestions at [email protected] or at [email protected]. Aditi Gupta Yajvender Pal Verma

Contents

Preface 1. Introduction 1.1 The Control System 1.2 Open Loop Control System 1.3 Closed Loop Control System 1.4 Basic Example 1.5 Classification of Control Systems 1.6 General Block Diagram of Closed Loop Control System 1.7 Illustrative Examples of Closed Loop Control System

v 1 1 1 2 3 4 5 7

2. Control System Representation 2.1 Introduction 2.2 Transfer Function Representation 2.3 Block Diagram Representation and Reduction 2.4 Signal Flow Graph Representation and Reduction

12 12 12 18 30

3. Modeling of Control Systems 3.1 Introduction 3.2 Modeling of Mechanical Systems 3.3 Modeling of Electrical Systems 3.4 Thermal Systems 3.5 Hydraulic Systems

48 48 48 51 58 60

4. Effects of Feedback on Control Systems 4.1 Introduction 4.2 Effect of Feedback on Output Due to Parameter Variations in Control System 4.3 Effect of Feedback on Sensitivity of Control System 4.4 Effect of Feedback on Gain of Control System 4.5 Effect of Feedback on Dynamics of Control System 4.6 Effect of Feedback on Stability of Control System 4.7 Effect of Feedback on Disturbances in Control Systems 4.8 Conclusions 4.9 Regenerative Feedback

65 65

65 66 71 72 73 73 77 77

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Contents

5. Time Response Analysis 5.1 Introduction 5.2 Important Terms 5.3 Standard Test Input Signals for Transient Analysis 5.4 Time Response of First Order System 5.5 Time Response of Second Order System 5.6 Time Response Specifications of Second Order System 5.7 Design Considerations: Performance Indices 5.8 Concept of Steady State Error 5.9 Effect of Changing Inputs on Steady State Error 5.10 Effect of G(s) H(s) or Type of System on Steady State Error 5.11 Summary of Steady State Errors and Error Constants

81 81 81 82 84 89 96 102 103 104 106 109

6. Concepts of Stability and Its Algebraic Solutions 6.1 Introduction 6.2 Concept of Stability 6.3 Types of Stability 6.4 Necessary Condition for Stability 6.5 Hurwitz Stability Criterion 6.6 Routh’s Stability Criterion 6.7 Special Cases of Routh’s Hurwitz Criterion 6.8 Applications of Routh’s Hurwitz Criterion 6.9 Relative Stability Analysis

123 123 123 127 127 128 129 131 137 141

7. Root 7.1 7.2 7.3 7.4 7.5 7.6 7.7

146 146 146 147 167 168 171 172

Locus Technique Introduction Concept of Root Locus: Map Criterion and Angle Criterion Construction Rules of Root Locus Determination of K for a Specified Damping Factor d Cancellation of Poles and Zeros Root Locus When m > n Root Contours

8. Frequency Response Analysis 8.1 CoRRelation Between Time Response and Frequency Response of a Second Order System 8.2 Frequency Response Specifications 8.3 Bode Plot 8.4 Polar Plot

179

179 182 183 200

Contents

8.5 Nyquist Stability Criterion 8.6 Non-Minimum Phase Systems, All Pass Systems, Minimum Phase Systems 9. Control System Components 9.1 Servomotors 9.2 Error Detectors 9.3 Tachogenerators 9.4 Stepper Motors

ix

208 223 231 231 237 241 243

10. Robust Control Systems 10.1 Introduction 10.2 System Sensitivity 10.3 Analysis of Robustness 10.4 PID Controllers 10.5 Tuning of PID Controllers

249 249 249 251 251 255

11. Sampled Data Control System 11.1 Introduction 11.2 Components of Sampled Data Control System 11.3 Advantages of Sampled Data Control System Over Analog Control System 11.4 Applications of Sampled Data Control System 11.5 Sampler 11.6 Z-transform 11.7 Hold Circuit/Signal Reconstruction 11.8 Pulse Transfer Function 11.9 Stability Analysis

260 260 261

Appendix I:

Laplace Transforms of Some Useful Functions

261 261 262 264 271 272 278 293

Answers to Unsolved Exercise (Selected Numericals)

295

References

297

Index

299

1

Introduction

CHAPTER

1.1

THE CONTROL SYSTEM

The control system is one of the most important aspects in engineering sciences. It also plays an important role in daily life cycle, in almost every facet of life. In the present world of increasing complexity and automation, more accurate, precise, safe and high productive systems are possible by using the concept of control engineering. The control system is defined as a system designed to manipulate or maintain some parameters of the operating system, according to our desire. A control system is an interconnection of some physical components that take the control action to achieve some desired output. There are three basic elements of a control system. These are input signal, system, and output signal. This is shown in Fig. 1.1.

Fig. 1.1

So, a control system presents the cause and effect relationship between input and output variables of a system. Some of the daily life examples of control systems are: (i) Heating rod: It heats the water i.e., it controls the temperature of the water. (ii) Flying a kite: The direction of the kite is controlled by its string. (iii) Switching a lamp: The ON and OFF of a lamp is controlled by its switch. (iv) Automatic washing machine: The running time of washing machine can be controlled manually by its switch.

1.2

OPEN LOOP CONTROL SYSTEM

These are the control systems in which output has no effect on control action. One output is obtained corresponding to one input. It is the system which has no feedback and output is not compared to reference input. The block diagram of an open loop control system is shown in Fig. 1.2.

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Automatic Control Engineering

Fig. 1.2

In this system, control signal is actuated by input signal only and is independent of output signal. To get the desired output or to change the output, input needs to be changed accordingly and hence the control signal. Controller can be amplifier, filter etc. Some day-to-day examples of open loop control systems are traffic light control, washing machine control, heating rod control and switching of a lamp. Advantages of open loop control systems: (i) These are simple systems as number of components is less. (ii) These are easily maintained and are economical. (iii) Their calibration is easy. (iv) They are more stable. Disadvantages of open loop control systems: (i) These are not precise and accurate. (ii) These are not reliable. (iii) Their optimisation is not possible.

1.3

CLOSED LOOP CONTROL SYSTEM

These are the control systems in which the control action depends upon the changes in the output. In these systems, output is continuously compared to reference input thereby generating a deviation or error signal that controls the control signal. As output is fed to the comparator for comparison to input, these systems are also called feedback control systems. The basic components of a closed loop control system are shown in Fig. 1.3.

Fig. 1.3

Some day-to-day examples of closed loop control systems are automatic electric iron, air conditioner and voltage stabilizer.

Introduction 3

Advantages of closed loop control systems: (i) These systems are more accurate and precise. (ii) These systems are more reliable. (iii) These systems are faster. (iv) These can be automated and optimized. Disadvantages of closed loop control systems: (i) Due to increase in the number of components, these systems are more complex. (ii) These are more costly. (iii) These are more unstable.

1.4

BASIC EXAMPLE

Driving system of an automobile: Figure 1.4 shows the open loop control system for an automobile. If the driver exerts a pressure on the accelerator, then a particular speed is obtained corresponding to that pressure through the engine of the vehicle. To change the speed or to get the desired speed, pressure on the accelerator needs to be changed. The output speed has no effect on the control action.

Fig. 1.4

Now, consider the closed loop control system for an automobile as shown in Fig 1.5, which is practically the case.

Fig. 1.5

In this system, output speed is continuously read by speedometer and is compared to a reference speed. Driver is continuously observing the deviation between reference speed and desired speed and is accordingly changing the pressure on the accelerator, so as to get the desired output. So, it is clear that control action taken by the driver is taken by firstly comparing the output to reference input. Thus, control signal is dependent upon output signal.

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Automatic Control Engineering

1.5

CLASSIFICATION OF CONTROL SYSTEMS

1.5.1 Manually Controlled and Automatic Control Systems The control system in which human operators are involved to perform the desired control action, are called manually controlled control systems. For example, the driving system of an automobile. When all the control actions are taken by some machines or equipment, the control systems are called automatic control systems. For example, temperature control system of a furnace, servomechanism.

1.5.2 Time Variant and Time Invariant Control Systems The control systems in which the parameters vary with time are called time variant control systems. For example, space vehicle system as its mass decreases with time due to fuel consumption. The control systems in which all the parameters do not vary with time are called time invariant control systems. For example, electric networks consisting of any combination of resistances, inductances and capacitances.

1.5.3 Linear and Non-linear Control Systems The control systems which obey the principle of superposition and homogeneity are called linear control systems. The principle of superposition and homogeneity states that the output obtained by the simultaneous application of number of inputs is equal to the resultant of all the outputs obtained by the application of individual inputs. No physical system in this world is a linear system. The control systems which do not obey the principle of superposition and homogeneity are called non-linear control systems. Almost all physical systems are non-linear but these non-linear systems are difficult to analyze. So, by making certain assumptions, these can be analyzed on linear basis.

1.5.4 Continuous Time and Discrete Time Control Systems If all the parameters of control systems are known for every instant of time, then these are called continuous time control systems. For example, position control of a dc motor. If some of the parameters of control systems are known for discrete instants of time, then these are called discrete time control systems. For example, all computer-based systems or microprocessor-based systems.

1.5.5 Lumped Parameter and Distributed Parameter Control Systems If control systems can be represented by normal differential equations, then these systems are called lumped parameter control systems. These systems consist of lumped components like electric networks of resistances, inductances and capacitances. If control systems can be represented by partial differential equations, then these systems are called distributed parameter control system. In these systems, parameters are distributed along the whole system. For example, a transmission line.

Introduction 5

1.5.6 Single Input-single Output Systems (SISO) and Multiple Input-Multiple Output Systems (MIMO) SISO system is shown in Fig. 1.6(a) and MIMO system is shown in Fig. 1.6(b).

Fig. 1.6

1.5.7 Deterministic and Stochastic Control Systems When the response of a control system is predictable and repeatable, then it is called deterministic control system. For example, electrical network. In this network, if some voltage is supplied, current starts flowing which is predictable as well as repeatable. When the response of a control system is neither predictable nor repeatable, then it is called stochastic control system. For example, position of shaft in a tracking system.

1.5.8 Open Loop and Closed Loop Control Systems Open loop control systems are non-feedback control systems while closed loop control systems are feedback control systems. These are already discussed in detail in Sections 1.2 and 1.3 respectively.

1.6

GENERAL BLOCK DIAGRAM OF CLOSED LOOP CONTROL SYSTEM

The general block diagram of a closed loop control system is shown in Fig. 1.7. The various components in the block diagram are error detector, control elements, feedback element and controlled system or plant. The various signals in the block diagram are input or reference signal, error or actuating signal, control signal, feedback signal and actual output signal. These are explained below:

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Automatic Control Engineering

(i) Input signal or reference signal: It is also known as desired output. This is the reference signal that is sent into the error detector. The aim of the control system is to make actual output equal to this reference signal. (ii) Error detector: It compares the actual output signal to the reference signal and produces the difference between the two, known as error signal or actuating signal. Potentiometer is one of the examples of error detector. (iii) Error signal: It is the difference between reference signal and actual output signal and thus is responsible for taking the control action by the control elements. (iv) Control elements: Sometimes, the level of error signal is too small to drive the control elements to take the control action. So, control elements increase the power level of error signal and then take the required control action. Amplifier is one of the examples of control elements. (v) Control signal: The signal that is produced by control elements and is applied to the operating system or plant to control or maintain its some parameter. (vi) Actual output signal: The actual output produced by the system or plant. It is also called response signal. (vii) Feedback element: It performs the following functions: • senses the output signal • converts the output signal into suitable form as that of reference input signal so that it can be compared to reference input signal. Transducer is one of the examples of feedback elements. (viii) Feedback signal: Sometimes, actual output signal is not in the same form as that of reference input signal, so it becomes difficult to compare them. Thus, feedback element is used that converts the output signal into a suitable form, known as feedback signal, so that it can be compared to reference signal for further processing. (ix) Controlled system or plant: It is the operating system whose parameter is to be maintained, controlled or changed by control system. (x) Controller: It is the combination of error detector and control elements.

Fig. 1.7

Introduction 7

1.7

ILLUSTRATIVE EXAMPLES OF CLOSED LOOP CONTROL SYSTEM

1.7.1 Servomechanism It is defined as the mechanism that is used to control the position of some operating system. ‘Servo’ is the technical name for ‘position’. So, servomechanism is the technique adopted for position control systems or the systems having output as the time derivative of position i.e., speed and acceleration. Hence, all the control systems which are used to control the position, speed or acceleration of some operating system involve servomechanism. One such example is “controlling the position of load shaft”. The control system designed for this is shown in Fig. 1.8.

Fig. 1.8

The output position of load shaft qc is fed to the potentiometer acting as error detector. The reference position is qr. qc and qr are compared by potentiometer and error signal VE is generated. This VE is proportional to (qr – qc ). The power level of VE is increased using amplifier. This increased VE then starts the DC servomotor to rotate in a particular direction so as to make the position of load shaft same as that of qr through gear system. DC servomotor goes on changing the position of load shaft as long as qc differs from qr and VE has some finite value. As soon as qc becomes equal to qr , VE becomes zero, so is the input to the servomotor and servomotor stops. So, no control action is taken as required. Various signals and elements are: Error detector : Potentiometer Controlled system : Load shaft Control elements : Amplifier, DC servomotor, gear system Reference signal : qr Actual output signal : qc Error signal : VE

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Automatic Control Engineering

1.7.2 Water Level Control in a Tank Output signal is height of water at any moment i.e., h. It is sensed by float and fed at one arm of the error detector (potentiometer). Required or reference water level is H which is fed to the second arm of the error detector. If h deviates from H, then error signal VE is generated which is further amplified by using amplifier. This amplified signal then drives the dc motor to control the opening and closing of inlet valve V2 and outlet valve V1. This process continues until h becomes equal to H. As soon as h becomes equal to H, VE becomes zero and DC motor stops thereby causing no control action. The control system is shown in Fig. 1.9. Various signals and elements are: Error detector : Potentiometer Controlled system : Water tank Control elements : Amplifier, DC motor Feedback element : Float Reference signal : H at A Actual output signal : h at B Error signal : VE Control signal : a V2

a

H

A

B

h

Pot. DC servomotor

Float H

h

Ampli Ampli er VE

V1

Fig 1.9

1.7.3 Temperature Control System The control system designed for controlling the temperature of an electric furnace is shown in Fig. 1.10. The output signal of electric furnace i.e., temperature is sensed by the thermocouple and is fed to A/D converter. Conversion of analog to digital signal is required due to the digital nature of electric comparator, where it is compared to the reference temperature r. Error signal e is generated which is proportional to (r – b), where b is the feedback signal. Now, this digital signal is again converted into analog signal using D/A converter. This error signal then operates the relay to cause ON or OFF of the heater. The output of heater further controls the temperature of the electric

Introduction 9

furnace. This process continues until b becomes equal to r. When b = r, no error signal is generated and relay will not operate as required.

Fig. 1.10

Various signals and elements are: Reference signal : r Actual output signal : Temperature(c) Error signal : e Feedback signal : b Error detector : Electric comparator Controlled system : Electric furnace Control elements : D/A converter, relay, heater Feedback elements : Thermocouple, A/D converter

Exercise 1. Explain with a neat diagram, the general block diagram of an automatic control system. 2. Differentiate between (i) Linear time variant and linear time invariant control systems. (ii) Continuous time and discrete time control systems. (iii) SISO and MIMO control systems. (iv) Linear and non-linear control systems. (v) Deterministic and stochastic control systems. (vi) Lumped parameter and distributed parameter control systems. 3. Compare open loop control systems and closed loop control systems. 4. Differentiate between open loop control systems and closed loop control systems with illustrative examples. 5. Explain the concept of servomechanism.

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Automatic Control Engineering

Objective Type Questions 1. Comparison of output signal and reference signal is done by (a) error detector (b) control elements (c) feedback element (d) sensor 2. Principle of superposition and homogeneity is followed by (a) non-linear time variant system (b) non-linear time invariant system (c) linear time invariant system (d) linear time variant system 3. Automatic washing machine is an example of (a) open loop control system (b) single feedback control system (c) multi-feedback control system (d) multi-variable closed loop system 4. Servomechanism controls the following parameters of the system (a) position (b) velocity (c) acceleration (d) all of these 5. Open loop control system is the one in which (a) input and output have effect on control action (b) output has effect on control action (c) output has no effect on control action (d) both (a) and (b) 6. Human system is an example of (a) open loop control system (b) single variable closed loop system (c) multi-variable closed loop system (d) none of these 7. Which of the following is an example of linear system? (a) y = a0x (b) y = a0x + a1 (c) y = a0 (d) y = a0x2 + a1x + a2 8. Air conditioner is an example of (a) open loop control system (b) single feedback control system (c) multi-feedback control system (d) none of these 9. Automatic iron is an example of (a) open loop control system (b) single feedback control system (c) multi-feedback control system (d) none of these 10. Which of the following systems have maximum stability? (a) open loop control system (b) single feedback control system (c) multi-feedback control system (d) none of these 11. In comparison to closed loop control system, open loop control system is (a) more stable, more accurate (b) more stable, less accurate (c) less stable, more accurate (d) less stable, less accurate

Introduction 11

12. Human eye is an example of (a) open loop control system (c) multi-feedback control system

(b) single feedback control system (d) none of these

Answers 1. (a) 6. (c) 11. (b)

2. (c) 7. (a) 12. (c)

3. (a) 8. (c)

4. (d) 9. (b)

5. (c) 10. (a)

2

Control System Representation

CHAPTER

2.1

INTRODUCTION

In previous chapter, physical control systems were discussed like servosystems, water level control in a tank, temperature control system, etc. These physical systems were basically the collection of physical objects connected together to fulfill an objective. These physical systems can also be represented mathematically in the following forms: (a) Transfer function representation (b) Block diagram representation (c) Signal flow graph representation

2.2

TRANSFER FUNCTION REPRESENTATION

Transfer function of a linear time invariant (LTI) system is defined as the ratio of Laplace transform of input variable of the system to the Laplace transform of output variable of the system, with the assumption that all the initial conditions are kept at zero. Consider a block diagram representation of the simplest open loop system: For Fig. 2.1, C (s) R (s) T (s) = G (s) = open loop transfer function C (s) = Laplace transform of output variable R (s) = Laplace transform of input variable

T (s) = G (s) =

where,

Fig. 2.1

Transfer function gives only the relationship between input and output variables of the system and does not tell anything about the internal state of the system.

Control System Representation

13

Consider an LTI system having input variable r (t) and output variable c (t), the input-output relationship can be described using nth order differential equation as shown: yn

d n c(t ) d n − 1 c (t ) dc(t ) d m r (t ) dr (t ) + + … + + = + … + x1 + x 0 r (t ) y y y c ( t ) x n−1 1 0 m dt n dt dt m dt dt n − 1

(2.1)

where x and y are constants. Now, take Laplace transform on both sides of equation (2.1),

( y n s n + y n - 1s n - 1 + ..... + y 1s + y 0 ) C (s ) = ( xm s m + xm - 1s m - 1 + .....x1s + x0 ) R (s ) fi

xm s m + xm − 1 s m − 1 + … x1 s + x0 C (s) = T(s) = G (s) = y sn + y sn − 1 + … + y s + y n n−1 1 0 R (s)

(2.2)

Equation 2.2 gives the mathematical representation of a control system in terms of transfer function. When m > n, transfer function is called improper. m = n, transfer function is called proper. m < n, transfer function is called strictly proper.

2.2.1 Various Terms Related to Transfer Function Following conclusions are drawn from equation (2.2) and accordingly the following terms are defined in relation to transfer function: (i) Characteristic equation: When denominator polynomial of transfer function is equated to zero, then characteristic equation is obtained i.e., y n sn + y n − 1 sn − 1 + … + y 1 s + y 0 = 0. (ii) Order of the system: The highest power of complex variable s in denominator of transfer function given in equation (2.2) gives the order of the system. n is the order of the system given by equation (2.2). (iii) Poles and zeros of a transfer function: The numerator and denominator polynomials in transfer function given by equation (2.2) can be written as the product of m and n linear factors respectively, as:

T (s) = G (s) = K

(s − z1 ) (s − z2 ) … (s − zm ) (s − p1 ) (s − p2 ) … (s − zn )

(2.3)

where K is called gain factor or scalar factor of transfer function. The values of complex variable s which make the transfer function zero, are called zeros of the transfer function. So, z1, z2, º, zm are zeros of the transfer function. These are represented by (small circle) on pole-zero diagram. The values of complex variable s which make the transfer function tend to infinity, are called

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Automatic Control Engineering

poles of the transfer function. So, p1, p2, º, pn are poles of the transfer function. Poles can be found out by solving characteristic equation for variable s. They are represented by × (small cross) on pole-zero diagram. (iv) Type of the system: The number of poles that lie on origin i.e., the number of values of s equal to zero in characteristic equation gives the type of the system. If either poles or zeros coincide, then they are called multiple or repeated poles or multiple or repeated zeros respectively, otherwise, they are called simple poles or simple zeros.

Example 2.1

For the transfer function given: s+5 T (s) = G (s) = 2 s (s + 6) (s 2 + 25) Find characteristics equation, pole-zero diagram, order and type of the system.

Solution: Step 1: Characteristics equation is:

(

)

(

s 2 ( s + 6 ) s 2 + 25 = 0 ⇒ s 3 + 6 s 2

)(s

2

)

+ 25 = 0 ⇒ s 5 + 6 s 4 + 25 s 3 + 150 s 2 = 0

Fig. 2.2

Step 2: Poles of the transfer function are at s = 0, 0, –6, ± 5 j. Step 3: Zero of the transfer function is at s = – 5. Step 4: Pole-zero diagram is shown in Fig. 2.2. Step 5: Order of the system is 5. Step 6: Type of the system is 2.

2.2.2 Transfer Function of Electric Circuits Number of examples using different electric circuits are solved:

Control System Representation

Example 2.2

15

Obtain the transfer function of the electric network shown in Fig. 2.3.

Fig. 2.3

Solution: Step 1: Applying Kirchhoff’s voltage law in input circuit, di (t ) 1 i(t ) dt − L =0 ∫ C dt Applying KVL in the output circuit,

(2.4)

vi (t ) − Ri (t ) −

Step 2:

di (t ) =0 dt Taking Laplace transform on both sides of equations (2.4) and (2.5), 1 I (s) Vi (s ) − RI (s ) − − LsI (s ) = 0 C s

fi

vo ( t ) − L

(2.5)

⎡ RCs + s 2 LC + 1 ⎤ Vi (s ) = I (s ) ⎢ ⎥ Cs ⎣ ⎦

(2.6)

and

Vo (s) – LsI (s) = 0

fi

Vo (s) = LsI (s)

(2.7)

Step 3: Dividing equation (2.7) by (2.6), Transfer function = T (s) =

Example 2.3

Vo (s ) = Vi (s )

I (s ) Ls ⎡ RCs + s 2 LC + 1 ⎤ I (s) ⎢ ⎥ Cs ⎣ ⎦

=

s 2 LC (Answer) s LC + sRC + 1 2

Obtain the transfer function of the system given in Fig. 2.4.

Fig. 2.4

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Solution: Step 1: Applying KVL in loop 1, vi (t ) − i1 (t ) − 3∫ [i1 (t ) − i2 (t )] dt = 0

Taking Laplace transform on both sides, Vi (s ) − I 1 (s ) − 3

I 1 (s) I (s) +3 2 =0 s s

(2.8)

Step 2: Applying KVL in loop 2, di2 (t ) − 3∫ [ i2 (t ) − i1 (t )] dt = 0 dt Taking Laplace transform on both sides, − 5i2 (t ) −

− 5 I 2 (s ) − sI 2 (s ) − 3

fi

I 1 (s) = I 2 (s)

s 3

I 2 (s) I (s) +3 1 =0 s s

⎡ ⎢⎣ 5 + s +

⎡ s 2 5s ⎤ 3⎤ = I ( s ) 2 ⎢ + + 1⎥ ⎥ s⎦ ⎣3 3 ⎦

(2.9)

Step 3: Also, in loop 2, di2 (t ) dt Taking Laplace transform on both sides, vo ( t ) = 1

Vo (s) = sI2 (s)

(2.10)

Step 4: Putting equation (2.9) in (2.8), Vi (s ) =

⎡ s 2 5s ⎤⎛ − 3I 2 (s) + I 2 ( s ) ⎢ + + 1⎥ ⎜ 1 + s ⎣3 3 ⎦⎝

3⎞ ⎟ s⎠

⎡ 3 s 2 5s 3⎤ = I 2 (s) ⎢ − + + + 1 + s + 5 + ⎥ s⎦ ⎣ s 3 3 ⎡ s 2 8s ⎤ = I 2 (s) ⎢ + + 6⎥ ⎣3 3 ⎦

(2.11)

Step 5: Dividing equation (2.10) by (2.11), Vo (s ) s 3s = 2 = 2 8s Vi (s ) s s + 8s + 18 + +6 3 3

(Answer)

Control System Representation

Example 2.4

17

Obtain transfer function of the circuit given in Fig. 2.5.

Fig. 2.5

Solution: Step 1: Applying KVL in loop 1, ei (t ) − R1 i (t ) − R2 i (t ) − L

Applying Laplace transform on both sides,

di(t ) =0 dt

Ei (s ) − R1 I (s ) − R2 I (s ) − LsI (s ) = 0 Ei (s ) = I (s ) [ R1 + R2 + sL ]

fi

(2.12)

Step 2: In loop 2, di(t ) dt Taking Laplace transform on both sides, Eo (s ) = R2 I (s ) + sLI (s ) eo (t ) = R2 i (t ) + L

fi

Eo (s ) = I (s ) [ R2 + sL ]

(2.13)

Step 3: Dividing equation (2.13) by (2.12), Eo (s ) R2 + sL = Ei (s ) R1 + R2 + sL

(Answer)

2.2.3 Transfer Function of a Closed Loop System For a negative feedback control system, shown in Fig. 2.6, the transfer function or the relationship between input and output variables is derived as follows:

Fig. 2.6

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The terms used in Fig. 2.6 are given below: R(s) = Reference Input Signal E (s) = Actuating or Error Signal C (s) = Output Signal B (s) = Feedback Signal G (s) = Open Loop Transfer Function or Forward Path Transfer Function = C (s)/E (s) H (s) = Feedback Path Transfer Function = B (s)/C (s) T (s) = Overall Transfer Function of the System or Closed Loop Transfer Function = C (s)/R (s) From Fig. 2.6, E (s) = R (s) – B (s) (2.14) = R (s) – C (s) H (s) and C (s) = E (s) G (s) (2.15) Putting (2.14) in (2.15), C (s) = [R (s) – C (s) H (s)] G (s) fi C (s) [1 + H (s) G (s)] = R (s) G (s) C (s) G(s ) = T (s) = R(s ) 1 + H (s ) G(s )

fi

(2.16)

Equation (2.16) gives the transfer function of a closed loop system. The transfer function obtained in equation (2.16) is obtained for negative feedback system, also known as degenerative system. If the output of system is fed back with positive sign, then transfer function is changed to C (s) G(s ) = T (s) = R(s ) 1 − H (s ) G(s )

and the system is called regenerative system. In general, T (s ) =

2.3

G(s ) , 1 ± H (s ) G(s )

+ sign for negative feedback – sign for positive feedback

BLOCK DIAGRAM REPRESENTATION AND REDUCTION

Another way of representing a physical control system mathematically is block diagram representation. As discussed earlier, the input-output behavior of a linear system is described by its transfer function as, T (s) = C (s)/R (s), where C (s) = Laplace transform of output variable R (s) = Laplace transform of input variable

Control System Representation

19

A convenient way to pictorially represent this transfer function is block diagram representation. It consists of one or more blocks and flow of signal. It is shown in Fig. 2.7.

Fig. 2.7

It is the pictorial representation of the functions performed by various components of physical systems (blocks) and flow of signal, which is unidirectional from input signal R (s) to output signal C (s), and is represented by arrows. The more complex systems comprising number of components are represented by interconnection of number of blocks for each element, connected by unidirectional signals. In addition to these blocks and signals, some more symbols are used in complex systems like summing or differencing point and take-off point as shown in Figs. 2.8 (a), (b) and (c). P

R

+

+

P

+

R –

Q

Q

(a) Summing point

(b) Differencing point

R=P+Q

R=P–Q

(c) Take-off point

Fig. 2.8

Block is the short hand pictorial representation of the cause and effect relationship between input and output of the system. Block diagrams of some control systems are very complex and require reduction or simplification to evaluate their performance. It can be obtained by block diagram reduction technique. The following rules are helpful in reducing a block diagram to find its transfer function: Rule 1: Combining Blocks in Cascade When two or more blocks are in cascade, the resultant block is product of two block’s transfer function as shown in Fig. 2.9.

Fig. 2.9

Rule 2: Combining Blocks in Parallel When two or more blocks are in parallel, the resultant block is sum of two block’s transfer function as shown in Fig. 2.10.

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Fig. 2.10

Rule 3: Moving a Take-off Point After the Block If take-off point is to be moved after the block, then the block with reciprocal of original transfer function appears in the signal of take-off point, as shown in Fig. 2.11.

Fig. 2.11

Rule 4: Moving a Take-off Point Before the Block If take-off point is to be moved before the block, then the block with same transfer function appears in the signal of take-off point, as shown in Fig. 2.12.

Fig. 2.12

Rule 5: Moving a Summing Point After the Block If summing point is to be moved after the block, then the block with same transfer function appears in the signal of summing point, as shown in Fig. 2.13.

Fig. 2.13

Control System Representation

21

Rule 6: Moving a Summing Point Before the Block If summing point is to be moved before the block, then the block with reciprocal of original transfer function appears in the signal of summing point, as shown in Fig. 2.14.

Fig. 2.14

Rule 7: Eliminating a Feedback Loop The formula derived in section 2.2.3 is used to eliminate the feedback loop, as shown in Fig. 2.15.

Fig. 2.15

Summary Rule Combining blocks in cascade Combining blocks in parallel

Moving a take-off point after the block Moving a take-off point before the block

Original Block Diagram

Equivalent Block Diagram

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Automatic Control Engineering

Moving a summing point after the block

Moving a summing point before the block

Eliminating a feedback loop

Example 2.5

Find transfer function of the following block diagram:

Fig. 2.16

Solution: Step 1: Moving take-off point after the block G3, Fig. 2.16 reduces to Fig. 2.17.

Fig. 2.17

Control System Representation

Step 2: Eliminating feedback loop, Fig. 2.17 reduces to Fig. 2.18.

Fig. 2.18

Step 3: Combining blocks in cascade, Fig. 2.18 reduces to Fig. 2.19. H3/G3 – R +

G1 –

G2 G3/(1 + G3H2)

+

H1

Fig. 2.19

Step 4: Eliminating feedback loop, Fig. 2.19 reduces to Fig. 2.20.

Fig. 2.20

Step 5: Combining blocks in cascade, Fig. 2.20 reduces to Fig. 2.21.

Fig. 2.21

Step 6: Eliminating feedback loop, Fig. 2.21 reduces to Fig. 2.22.

Fig. 2.22

C

23

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So, transfer function becomes, T (s) = Example 2.6

G1 G2 G3 1 + G2 H 3 + G3 H 2 + G1 G2 G3 H 1

(Answer)

Obtain transfer function of the following block diagram:

Fig. 2.23

Solution: Step 1: Moving take-off point between G3 and G4 after G4 and summing point between G1 and G2 before G1, Fig. 2.23 reduces to Fig. 2.24.

Fig. 2.24

Step 2: Combining first and second summing points, Fig. 2.24 reduces to Fig. 2.25.

Fig. 2.25

Step 3: Simplifying first and second feedback loops, Fig. 2.25 reduces to Fig. 2.26.

Fig. 2.26

Control System Representation

25

Step 4: Combining the blocks in cascade, Fig. 2.26 reduces to Fig. 2.27.

Fig. 2.27

Step 5: Eliminating feedback loop, Fig. 2.27 reduces to Fig. 2.28. R

G1G2G3G4/(1 + G1G2H1 + G3G4H2 + G1G2G3G4H1H2 – G2G3H3)

C

Fig. 2.28

So, transfer function is, T (s) =

G1 G2 G3 G4 1 + G1 G2 H 1 + G3 G4 H 2 + G1 G2 G3 G4 H 1 H 2 − G2 G3 H 3

(Answer)

Example 2.7 Obtain the transfer functions C/R and C/D of the system shown in Fig. 2.29.

Fig. 2.29

Solution: Step 1: To find C/R, second input of the system i.e., D is put equal to zero, so that Fig. 2.29 reduces to Fig. 2.30.

Fig. 2.30

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Automatic Control Engineering

Step 2: Combining G1 and GP in cascade and moving summing point between GC and G1 before GC, Fig. 2.30 reduces to Fig. 2.31.

Fig. 2.31

Step 3: Combining two summing points and combining GC and G1 GP in cascade, Fig. 2.31 reduces to Fig. 2.32.

Fig. 2.32

Step 4: Applying parallel blocks and feedback blocks rules, Fig. 2.32 reduces to Fig. 2.33.

Fig. 2.33

Step 5: Combining blocks in cascade, Fig. 2.33 reduces to Fig. 2.34. R

(GC + Gf ) + (G1GP) (1 + G1 GC GP H)

C

Fig. 2.34

(

)

     GC + G f (G1 GP ) So, C = R 1 + G1 GC GP H

(Answer 1)

Step 6: To find C/D, first input of the system i.e., R is put equal to zero, so that Fig. 2.29 reduces to Fig. 2.35.

Control System Representation

27

Fig. 2.35

Step 7: Combining G1, GC, and –H in cascade, Fig. 2.35 reduces to Fig. 2.36.

Fig. 2.36

Step 8: Eliminating feedback loop, Fig. 2.36 reduces to Fig. 2.37.

Fig. 2.37

So,

    GP C = D 1 + G1 GP GC H

Example 2.8 in Fig. 2.38.

(Answer 2)

Obtain all the possible transfer functions for the MIMO system shown

Fig. 2.38

Solution: Step 1: There are four possible transfer functions i.e., C1/R1, C1/R2, C2/R1 and C2/R2.

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Step 2: To find C1/R1, put C2 = R2 = 0, Fig. 2.38 reduces to 2.39.

Fig. 2.39

Step 3: G2, G3 and G4 are in cascade, so Fig. 2.39 reduces to Fig. 2.40.

Fig. 2.40

Step 4: Eliminating feedback loop, Fig. 2.40 reduces to Fig. 2.41.

Fig. 2.41

Step 5: To find C1/R2, put C2 = R1 = 0, Fig. 2.38 reduces to 2.42.

Fig. 2.42

Control System Representation

Step 6: G2, G3 and G4 are in cascade, Fig. 2.42 reduces to Fig. 2.43.

Fig. 2.43

Step 7: Eliminating feedback loop, Fig. 2.43 reduces to Fig. 2.44.

Fig. 2.44

Step 8: To find C2/R1, put C1 = R2 = 0, Fig. 2.38 reduces to 2.45.

Fig. 2.45

Step 9: G1, G2 and G4 are in cascade, Fig. 2.45 reduces to Fig. 2.46.

Fig. 2.46

Step 10: Eliminating feedback loop, Fig. 2.46 reduces to Fig. 2.47.

Fig. 2.47

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Step 11: To find C2/R2, put C1 = R1 = 0, Fig. 2.38 reduces to Fig. 2.48.

Fig. 2.48

Step 12: G1, G2 and G3 are in cascade, Fig. 2.48 reduces to Fig. 2.49.

Fig. 2.49

Step 13: Eliminating feedback loop, Fig. 2.49 reduces to Fig. 2.50.

Fig. 2.50

Step 14: The transfer functions are:     −G1G3 G4 C1     G1 C , 1 = = R1 1 − G1 G2 G3 G4 R2 1 − G1G2 G3 G4

    − G1 G2 G4 C 2 C2     G4 = , = R2 1 − G1 G2 G3 G4 R2 1 − G1 G2 G3 G4

2.4

(Answer)

SIGNAL FLOW GRAPH REPRESENTATION AND REDUCTION

Block diagram representation is a very advantageous approach to represent a control system. But it has one disadvantage that, for complicated systems, it becomes very

Control System Representation

31

difficult and sometimes impossible to reduce the block diagrams. A new and alternate method to represent a control system is signal flow graph (SFG) representation, which was developed by S.J. Mason. A readymade gain formula is available in this representation that relates the system input and output variables and directly gives the transfer function of the system without any transformation. An SFG representation is a graphical figure that represents a set of linear equations, giving relationship between various variables of the system. It basically consists of two basic elements: node and branch. Node is a small circle representing a variable of the system. Branch is a line connecting two nodes and represents the flow of signal. Branch always contains one arrow, giving the direction of flow of signal.

2.4.1 Construction of SFG from Set of Linear Equations Consider a set of linear equations: x2 = h12 x1 + h42 x4 x3 = h13 x1+ h23 x2 + h33 x3 x4 = h24 x4+ h34 x4

(2.17)

First step is to represent the various variables i.e., x1, x2, x3 and x4 by nodes, shown in Fig. 2.51(a). The first equation in (2.17) states that x2 is sum of two signals, coming from x1 and x4 and its SFG is shown in Fig. 2.51(b).

Fig. 2.51

Similarly, SFG for the remaining equations are shown in Figs. 2.51 (c) and (d) and complete SFG is shown in Fig. 2.51(e).

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2.4.2 Construction of SFG From Block Diagram There are some rules which are to be kept in mind while converting a block diagram into SFG: (i) Every signal in block diagram becomes a node in SFG. (ii) Every block in block diagram becomes a signal in SFG. (iii) Every summing point and take-off point in block diagram becomes a node in SFG. (iv) If take-off point comes just after a summing point, then only one node is made for both. Consider a block diagram shown in Fig. 2.52

Fig. 2.52

Two nodes i.e., R and C are for two signals R and C. Node x1 is for summing point S1 and for take-off point T1 as T1 is coming just after S1.

Fig. 2.53

x2 is for summing point S2. x3 is for take-off point T2. x4 is for summing point S3 and for take-off point T3. Nodes for T2 and S3 are not combined, as T2 is coming before S3. The complete signal flow graph is shown in Fig. 2.53.

2.4.3 Important Terminology for Signal Flow Graph In addition to nodes and branches, some more terms are useful to study the gain formula. These terms are defined below:

Control System Representation

33

(i) Input node or source node: Node representing the input variable in block diagram or node having only outgoing branches, e.g., node R in Fig. 2.53. (ii) Output node or sink node: Node representing the output variable in block diagram or node having only incoming branches, e.g., node C in Fig. 2.53. (iii) Path: It is the interconnection of branches in the direction of arrows, such that no node is followed more than once, e.g., R – x1 – x2 – x3, x1 – x2 – x3, R – x4 – C are some of the paths in Fig. 2.53. (iv) Forward path: It is the type of path starting from input node and terminating on output node, e.g., R – x1 – x2 – x3 – x4 – C, R – x1 – x4 – C are two forward paths in Fig. 2.53. (v) Forward path gain: It is the gain obtained by the product of all branch gains of a forward path, e.g., forward path gain for the forward path R – x1 – x2 – x3 – x4 – C is G1 G2 in Fig. 2.53. (vi) Loop: It is the special type of path, which starts and terminates at the same node. Except starting or terminating node, no node is to be followed more than once, e.g., x1 – x2 – x3 – x4 – x1, x1 – x4 – x1, x2 – x3 – x2, are the three loops in Fig. 2.53. (vii) Loop gain: It is the gain obtained by the product of all branch gains of a loop, e.g., loop gain for the loop x2 – x3 – x2 is – G2 H1 in Fig. 2.53. (viii) Non-touching loops: Two or more loops are non-touching, if they do not have any common node, e.g., x1 – x4 – x1 and x2 – x3 – x2 are two non-touching loops in Fig. 2.53.

2.4.4

Mason’s Gain Formula

It is the readymade gain formula giving the relationship between input and output variable. It is very useful in finding transfer function of a signal flow graph. It is given by: ∑ Pi Δ i T (s) =

i

Δ

where T (s) = overall transfer function of the system. Pi = forward path gain of ith forward path. Di = 1 – (sum of loop gains of those loops which are not touching ith forward path). D = Determinant of the graph = 1 – (sum of loop gains of all individual gains) + (sum of loop gain products of all possible combinations of two non-touching loops) – (sum of loop gain products of all possible combinations of three non-touching loops) +º Consider the signal flow graph of Fig. 2.53. Its overall transfer function can be found out using Mason’s gain formula. (a) There are two forward paths. Forward path gains are: P1 = G1 G2 P2 = G3

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Automatic Control Engineering

(b) There are three loops. Loop gains are:

L1 = – G1 G2 H2 L2 = – G2 H1 L3 = – G3 H2 (c) P1 touches all the loops. So, D1 = 1 – 0 = 1 P2 does not touch L2. So, D1 = 1 – L2 = 1 + G2 H1 (d) There is one possible combination of two non-touching loops i.e., L2 L3. So, D = 1 – (L1 + L2 + L3) + (L2 L3) = 1 + G1 G2 H2 + G2 H1 + G3 H2 + G2 G3 H1 H2 Now, the overall transfer function is given by:

C P1 Δ1 + P2 Δ 2 = R Δ G1 G2 + G3 (1 + G2 H 1 ) = 1 + G1 G2 H 2 + G2 H 1 + G3 H 2 + G2 G3 H 1 H 2

T (s) =

T (s) =

fi Example 2.9

G1 G2 + G3 + G2 G3 H 1 1 + G1 G2 H 2 + G2 H 1 + G3 H 2 + G2 G3 H 1 H 2

(Answer)

Obtain the transfer function of the signal flow given in Fig. 2.54.

Fig. 2.54

Solution: Step 1: There are three forward paths. Forward path gains are: P1 = G1 G2 G3 G4 G5; P2 = G1 G6 G4 G5 ; P3 = G1 G2 G7 Step 2: There are four loops. Loop gains are: L1 = – G4 H1; L2 = – G6 G4 G5 H2; L3 = – G2 G3 G4 G5 H2; L4 = – G2 G7 H2 Step 3: First and second forward paths touch all loops: Thus, D1 = 1 – 0 = 1 and

D2 = 1 – 0 = 1

Third forward path does not touch L1. So, D3 = 1 – L1 = 1 + G4 H1. Step 4: There is one pair of two non-touching loops i.e., L1 and L4 Thus, D = 1 – (L1 + L2 + L3 + L4) + (L1 L4) = 1 + G4 H1 + G6 G4 G5 H2 + G2 G3 G4 G5 H2 + G2 G7 H2 + G4 H1 G2 G7 H2

Control System Representation

35

Step 5: According to Mason’s gain formula, C (s ) P1 Δ1 + P2 Δ 2 + P3 Δ 3 + P4 Δ 4 Transfer function = R(s ) fi Transfer function = Example 2.10

G1 G2 G3 G4 G5 + G1 G4 G5 G6 + G1 G2 G7 (1 + G4 H 1 ) 1 + G4 H 1 + G4 G5 G6 H 2 + G2 G3 G4 G5 H 2 + G2 G7 H 2 + G2 G4 G7 H 1 H 2

(Answer) Find the transfer function of the signal flow graph shown in Fig. 2.55. G1

G3

G2

R

H1

G7

G8

G5

G4

C

H2 G6

H3

Fig. 2.55

Solution: Step 1: There is one forward path. Forward path gain is: P1 = G1 G2 G3 G4 G5 Step 2: There are four loops. Loop gains are: L1 = G2 H1; L2 = G4 H2; L3 = G7 H3; L4 = G1 G2 G3 G4 G5 G6 G7 G8 Step 3: First forward path does not touch L3. Thus,

D1 = 1 – L3 = 1 – G7 H3

Step 4: There are three pairs of two non-touching loops i.e., L1 L2, L2 L3 and L3 L1 and there is one pair of three non-touching loops i.e., L1 L2 L3. Thus, D = 1 – (L1 + L2 + L3 + L4) + (L1 L2 + L2 L3 + L3 L1) – (L1 L2 L3) = 1 – G2 H1 – G4 H2 – G7 H3 – G1 G2 G3 G4 G5 G6 G7 G8 + G2 G4 H1 H2 + G4 G7 H2 H3 + G2 G7 H1 H3 – G2 G4 G7 H1 H2 H3 Step 5: According to Mason’s gain formula, C (s ) P1 Δ1 = Transfer function = Δ R(s ) fi Transfer function = G1 G2 G3 G4 G5 ( 1 − G7 H 3 ) 1 − G2 H 1 − G4 H 2 − G7 H 3 − G1 G2 G3 G4 G5 G6 G7 G8 + G2 G4 H 1 H 2 + G4 G7 H 2 H 3 + G2 G7 H 1 H 3 – G2 G4 G7 H 1 H 2 H 3

(Answer) Example 2.11 Obtain the transfer function of the following block diagram. Convert it into signal flow graph and again obtain transfer function using Mason’s gain formula. Verify the results obtained.

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Fig. 2.56

Solution: Step 1: Block diagram reduction: Redraw the block diagram given in Fig. 2.56 to separate two feedback loops as in Fig. 2.57.

Fig. 2.57

Step 2: Eliminate feedback loop and cascade H1 and H2 to obtain Fig. 2.58.

Fig. 2.58

Step 3: Eliminate feedback loop to get Fig. 2.59.

Fig. 2.59

So, transfer function =

G1 G2 C = R 1 − G2 H 2 + G1 G2 H 1 H 2

Step 4: Signal flow graph:

Fig. 2.60

(Answer 1)

Control System Representation

37

Step 5: There is one forward path. Forward path gain is: P1 = G1 G2 Step 6: There are two loops. Loop gains are: L1 = G2 H2; L2 = – G1 G2 H1 H2 Step 7: First forward path touches all loops. So, D1 = 1 – 0 = 1 Step 8: There is no combination of non-touching loops. So, D = 1 – (L1 + L2) = 1 – G2 H2 + G1 G2 H1 H2 Step 9: According to Mason’s gain formula, Transfer function =

G1 G2 C P1 Δ 1 = = 1 − G2 H 2 + G1 G2 H 1 H 2 R Δ

(Answer 2)

Step 10: Answer 1 and Answer 2 are same. So, results are verified.

Exercise 1. Find characteristic equation, order of the system, type of the system and poles and zeros of the systems whose transfer functions are given by: 4 (s + 2) 10 (i) (ii) 2 2 s (s + 7 s + 12) s (1 + 0.4 s ) (1 + 0.1 s )

(iii)

16 (1 + 0.5 s ) s (1 + 0.125 s ) (1 + 0.1 s )

(iv)

2

80 (s + 5) s 3 (1 + 8 s )

2. A system is given by the block diagram shown in Fig. 2.61.

Fig. 2.61

Determine (i) Characteristic equation (ii) order (iii) type (iv) poles and zeros in s-plane. 3. A second order system is given by

d 2 z (t ) dz (t ) + 10 + 5 z (t ) dt dt Obtain transfer function, characteristic equation, pole and zero behavior of the system. y (t ) = 2

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4. Obtain the transfer function of the system shown in Fig. 2.62.

Fig. 2.62

5. Derive the transfer function of the electric circuit shown in Fig. 2.63.

Fig. 2.63

6. Obtain the transfer function of the system shown in Fig. 2.64.

Fig. 2.64

7. Obtain the transfer function of the system shown in Fig. 2.65 by block diagram reduction method and by signal flow graph method. H1 –

+ R

G1 –

G2

+

G3

G4

–

+

G6 +

H2

G5

Fig. 2.65

+

C

Control System Representation

39

8. Find the transfer function of the signal flow graph given in Fig. 2.66.

Fig. 2.66

9. Find the transfer function of the signal flow graph given in Fig. 2.67.

Fig. 2.67

10. Obtain the transfer function of the system shown in Fig. 2.68 by block diagram reduction method and by signal flow graph method.

Fig. 2.68

11. Explain the terms with a suitable example: path, forward path, forward path gain, loop, loop gain, non-touching loops.

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Objective Type Questions 1. Consider the following signal flow graphs. The value of gain is 2 for

(a) 2 and 3 (b) 1 (c) 3 (d) 2 2. Which of the following is open loop system? (a) The respiratory system of a man (b) A system for controlling the movement of the slide of a copying milling machine. (c) Traffic light control (d) A thermostatic control 3. The transfer function Y/R of the system shown is R

+

1/(s + 1)

Y

–

+ 1/(s + 1)

–

(a) 0 (b) 1/(s + 1) (c) 2/(s + 1) (d) 2/(s + 3) 4. The block diagram shown in Fig. 1 is equivalent to

Fig. 1

Control System Representation

2

2s + 6s + 5

41

5. The transfer function of the system is . The characteristic equation 2 ( s + 1) ( s + 2) is (a) 2 s2 + 6 s + 5 = 0 (b) (s + 1)2 (s + 2) = 0 (c) 2 s2 + 6 s + 5 + (s + 1)2 (s + 2) = 0 (d) 2s2 + 6 s + 5 – (s + 1)2 (s + 2) = 0 6. Transfer function of the integrator shown in Fig. 2 is

Fig. 2

(a) (c)

RCs RCs + 1 1 RC + s

1

(b) RCs + 1 (d)

7. Type-2 system has (a) two poles at origin (b) (c) three poles at origin (d) 8. Roots of characteristic equation is (a) zeros of the system (b) (c) poles of the system (d) 9. Which of the following transfer functions (s + 3) ( s + 7) (b) (a) (s + 6) (s + 8)

(c)

(s + 3) ( s + 7) s 2 (s + 6) (s + 8)

(d)

RC 1+s

one pole at origin two zeros at origin order of the system type of the system indicate type-0 system? (s + 3) ( s + 7) s (s + 6) (s + 8) (s + 3) ( s – 7) s 3 (s + 6) (s + 8)

10. The number of forward paths in the following signal flow graph are

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(a) 3 (c) 1

(b) 2 (d) 4

1 11. The open loop transfer function of unity feedback system is . The poles (2 + s )2 of the system are (a) 2 ± j (b) 1 ± 2 j (c) – 2 ± j (d) – 1 ± 2 j 12. The transfer function X2/X1 for the following signal flow graph is

(a) 1/(1 – T12)

(b) T12/(1 – T22)

(c) 1/(1 + T12) 13. X3/X1 is

(d) T12/(1 + T22)

(a) 20 (c) 5

(b) 50 (d) 0.5

Answers 1. (d) 6. (b) 11. (a)

2. (c) 7. (a) 12. (b)

3. (b) 8. (c) 13. (a)

4. (c) 9. (a)

5. (b) 10. (a)

MATLAB Programs P1. Find zeros, poles and gain of transfer function T (s) =

s2 + 4 s + 3 s + 2 s + 3 s3 + 5 s2 + 4 s + 1 5

4

Program: n = [1 4 3]; d = [1 2 3 5 4 1]; [z, p, k] = tf2zp(n,d)

% Numerator of transfer function % Denominator of transfer function % Conversion of transfer function to zeros, poles and gain

Control System Representation

43

Execution: z= –3 -1 p= 0.2267 + 1.4677i 0.2267 – 1.4677i -1.0000 + 0.0000i -1.0000 - 0.0000i -0.4534 k= 1 P2. Find zeros, poles and gain of transfer function T (s) =

10 (s + 1) (s + 2) (s + 4)

Program: n=conv([0 10],[1 1]); d=conv([1 2],[1 4]); [z, p, k] = tf2zp(n,d)

% Numerator of transfer function % Denominator of transfer function % Conversion of transfer function to zeros, poles and gain

Execution: z= -1 p= -4 -2 k= 10 P3. Find the transfer function of the control system whose zeros and poles are at – 2, – 1 and at – 4, – 2, – 5, – 6 respectively, having gain 10. Program: z=[-2;-1]; p=[-4;-2;-5;-6]; k=[10]; [n,d] = zp2tf(z, p, k) T=tf(n,d) Execution: n= 0 0 10 30 20

% Zeros of transfer function % Poles of transfer function % Gain of transfer function %Conversion of zeros, poles and gain into numerator and denominator of transfer function % Transfer function

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d= 1 17 104 268 240 Transfer function: 10 s^2 + 30 s + 20 -----------------------------------s^4 + 17 s^3 + 104 s^2 + 268 s + 240 P4. Plot pole-zero plot of the system having transfer function T =

s3 + 2 s2 + s + 3 s + 5 s3 + 3 s2 + 4 s + 1 4

Program: n=[1 2 1 3]; d=[1 5 3 4 1]; T=tf(n, d) pzmap(T)

% % % %

Numerator of transfer function Denominator of transfer function Transfer function Plot of pole-zero map

Execution: Transfer function: s^3 + 2 s^2 + s + 3 ----------------------------s^4 + 5 s^3 + 3 s^2 + 4 s + 1

Fig. F.1

P5. Find the overall transfer function for the system shown in Fig. F.2.

Fig. F.2

Control System Representation

Program: % To find G1 n1=[10]; d1=[1 2]; G1=tf(n1,d1); % To find G2 n2=[4]; d2=[1 1 1]; G2=tf(n2,d2); % To find overall transfer function T=series(G1,G2) Execution: Transfer function: 40 --------------------s^3 + 3 s^2 + 3 s + 2 P6. Find the overall transfer function for the system shown in Fig. F.3.

Fig. F.3

Program: % To find G1 n1=[40]; d1=[1 2 1]; G1=tf(n1,d1); % To find G2 n2=[2]; d2=[1 1]; G2=tf(n2,d2); % To find overall transfer function T=parallel (G1,G2)

45

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Execution: Transfer function: 2 s^2 + 44 s + 42 --------------------s^3 + 3 s^2 + 3 s + 1 P7. Find the overall transfer function for the system shown in Fig. F.4.

Fig. F.4

Program: % To find G n1=[3]; d1=[1 1 2]; G=tf(n1,d1); % To find H n2=[0.3]; d2=[1 1]; H=tf(n2,d2); % To find overall transfer function T=feedback(G,H) Execution: Transfer function: 3s+3 ----------------------s^3 + 2 s^2 + 3 s + 2.9 P8. Find the overall transfer function for the system shown in Fig. F.5.

Fig. F.5

Control System Representation

Program: % To find G1 n1=[10]; d1=[4 1]; G1=tf(n1,d1); % To find G2 n2=[2]; d2=[3 2]; G2=tf(n2,d2); % To find G3 G3=[1]; % To find H1 H1=[2]; % To find H2 H2=[5]; % To find overall transfer function feedback(series(parallel(G3,G1),feedback(G2,H1)),H2) Execution: Transfer function: 8 s + 22 ------------------12 s^2 + 67 s + 116

47

3

Modeling of Control Systems

CHAPTER

3.1

INTRODUCTION

The interconnection of various physical objects that has the aim to meet some objectives is called physical system. On the basis of types of physical components, these are further categorized as mechanical systems, electrical systems, hydraulic systems, thermal systems etc. One of the most and important steps in design and analysis of control systems is the equivalent representation of these physical control systems into mathematical models. The derivation of mathematical equations with the help of differential equations, for these physical control systems is called their mathematical modeling. These models describe the dynamic behavior of the systems and are further helpful in making the block diagrams and signal flow graphs of these systems. Since most of the control systems consist of mechanical, electrical or both types of components, so mathematical modeling of mechanical systems and electrical systems and their analogy to each other will be discussed in this chapter.

3.2

MODELING OF MECHANICAL SYSTEMS

Mechanical systems are further categorized on the basis of their motion as translational mechanical systems and rotational mechanical systems.

3.2.1 Translational Mechanical Systems These are the mechanical systems which move along a straight line or a fixed axis or a curved path. These systems are analyzed by using three idealized elements: mass, spring and damper. The input variable to these components is force (F) and output variable is displacement (x) or time derivative of displacement i.e., velocity (v) and acceleration (a). Mass: It is defined as the quantity of matter in a body and is concentrated at the centre of the body. It is a constant for a particular body. Figure 3.1 shows that when a mass of body M is applied with input signal force F, then a linear displacement (x) takes place along a straight axis.

Modeling of Control Systems 49

The relationship between these variables F and x is given by Newton’s law as in equation (3.1), d2 x (3.1) F=M dt 2 2 d x where a = , is the acceleration produced in the body in m/s2 dt 2

Fig. 3.1

Spring: It is a component of translational mechanical system that stores energy in the form of potential energy. Fig. 3.2 shows a linear spring of negligible mass, having spring constant K.

Fig. 3.2

The force-displacement relationship is given by Hook’s law as in equation (3.2), F = Kx (3.2) The units of spring constant K is N/m or pounds/ft. Damper or friction: Damper opposes the motion of mechanical elements. It can be coloumb frictional force, viscous frictional force or friction. Fig. 3.3 shows the damper having friction constant or damping constant f.

Fig. 3.3

The force-displacement relationship is given by equation (3.3), dx (3.3) F=f dt Now, consider the simplest translational mechanical system consisting of all the three idealized elements i.e., mass, spring and damper as shown in Fig. 3.4.

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Fig. 3.4

If the force F is exerted on this system, then a linear displacement x takes place, so that the input-output relationship becomes, d2x dx + f + Kx 2 dt dt Taking Laplace transform on both sides, F (s) = s2 MX (s) sf X(s) + KX (s)

F (t) = M

fi

X (s ) 1 = 2 F (s) s M + sf + K

(3.4) (3.5) (3.6)

Equation (3.6) gives the transfer function of the system shown in Fig. 3.4. This process of writing mathematical equations (here transfer function) of a physical system is known as modeling of system.

3.2.2 Rotational Mechanical System These are the mechanical systems which move around a fixed axis such that the displacement is angular displacement q. These systems are also analyzed using three idealized elements, viz., moment of inertia, spring and friction or damper. The input variable to these elements is torque and output variable is angular displacement q, or time derivative of q. Moment of inertia ( J ): The input-output relationship for the system shown in Fig. 3.5 is given by equation (3.7) T= J

d2 θ dt 2

(3.7) J T

q

Fig. 3.5

Spring: The input-output relationship for the system shown in Fig. 3.6 is given by equation (3.8) T = Kq (3.8)

Fig. 3.6

Modeling of Control Systems 51

Damper or friction: The input-output relationship for the system shown in Fig. 3.7 is given by equation (3.9) dθ (3.9) T=f dt

Fig. 3.7

Now, consider a simple rotational mechanical system as shown in Fig. 3.8, consisting of all the three idealized elements, viz., moment of inertia, spring and damper.

Fig. 3.8

If torque (T) is exerted on this system, then an angular displacement q takes place. The input-output relationship between T and q is given by equation (3.10) T= J

d2θ dθ + f + Kθ 2 dt dt

Taking Laplace transform on both sides, T = s2 Jq (s) + sf q (s) + Kq (s) θ (s) 1 = 2 fi T (s ) s J + sf + K

(3.10) (3.11) (3.12)

Equation (3.12) gives the transfer function of the system shown in Fig. 3.8.

3.3

MODELING OF ELECTRICAL SYSTEMS

The three basic components of all electrical systems are resistance, inductance and capacitance. If some voltage v is applied across these components, then current i flows through them. Thus, input-output relationship between v and i for these individual components are given by equations (3.13), (3.15) and (3.17).

fi

v (t) = Ri (t) v(t ) i (t) = R di (t ) v (t) = L dt

(3.13) (3.14) (3.15)

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1 v (t ) dt L∫ 1 i (t ) dt v (t) = C∫ dv (t ) i (t) = C dt

i (t) =

fi

fi

Fig. 3.9

3.3.1

(3.16) (3.17) (3.18)

Fig. 3.10

Fig. 3.11

Series R-L-C Circuit

Consider a series R-L-C circuit as shown in Fig. 3.12. Voltage v is applied to it, so that current i flows through the circuit. So, the input-output relationship between v and i is given by equation (3.19), using KVL, v (t) = Ri (t ) + L

di (t ) 1 + ∫ i (t ) dt dt C

(3.19)

Fig. 3.12

Putting i (t) =

dq (t ) i.e., rate of change of charge q (t), equation (3.19) becomes, dt dq(t ) d 2 q (t ) 1 R + L + q( t ) v (t) = (3.20) dt dt 2 C

Taking Laplace transform on both sides, 1 Q(s ) (3.21) V(s) = sRQ (s) + s2 LQ (s) + C Q(s ) 1 fi (3.22) = 1 V (s) s 2 L + sR + C Equation (3.22) gives the transfer function of the series R-L-C circuit shown in Fig. 3.12. Comparing equations (3.5), (3.11) and (3.21), the differential equations are found to be similar. Such systems whose differential equations are of similar or identical form are called analogous systems.

Modeling of Control Systems 53

From equations (3.5) and (3.21), force-voltage analogy is obtained. From equations (3.11) and (3.21), torque-voltage analogy is obtained. These analogies are shown in Table 3.1. Table 3.1 Translational Mechanical System

Rotational Mechanical System

Electrical System (Series R-L-C)

Force (F)

Torque (T)

Voltage (V)

Linear displacement (x)

Angular displacement (q)

Charge (q)

Mass (M)

Moment of inertia (J)

Inductance (L)

Friction (f )

Friction (f )

Resistance (R)

Spring (K)

Spring (K)

Reciprocal of capacitance (1/C)

3.3.2 Parallel R-L-C Circuit Consider a parallel R-L-C circuit as shown in Fig. 3.13. A current source I is connected across it, so that voltage drop v takes place across each element. Applying KCL, the input-output relationship between I and v becomes, v (t ) 1 dv (t ) + ∫ v (t ) dt + C (3.23) I (t) = R L dt

Fig. 3.13

Putting v (t) =

dφ ( t ) i.e., rate of change of flux f (t), equation (3.23) becomes, dt d 2 φ (t ) 1 dφ (t ) 1 + φ t + C ( ) I (t) = (3.24) R dt L dt 2

Taking Laplace transform on both sides, I (s) = s φ (s ) = I (s )

1 1 φ (s ) + φ (s ) + s 2 C φ (s ) R L 1

(3.25)

(3.26) 1 1 + R L Equation (3.26) gives the transfer function of the system shown in Fig. 3.13. Comparing equations (3.5), (3.11) and (3.25), force/torque-current analogy can be obtained as given in Table 3.2. fi

s2 C + s

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Table 3.2 Translational Mechanical System Force (F)

Rotational Mechanical System Torque (T)

Electrical System (Parallel R-L-C) Current (I)

Linear displacement (x)

Angular displacement (q)

Flux (f)

Mass (M)

Moment of inertia (J)

Capacitance (C)

Friction (f )

Friction (f )

Reciprocal of resistance (1/R)

Spring (K)

Spring (K)

Reciprocal of inductance (1/L)

The concept of analogous system is very useful because if the solution of only one system is obtained, then it can be extended to all other systems that are analogous to it. Normally, it is easier to express non-electrical systems in terms of electrical systems because electrical systems can be studied and analyzed easily through experiments. Example 3.1 Determine the system equations for the translational mechanical system as shown in Fig. 3.14. Also, draw force-voltage and force-current analogous systems.

f2

K2

2 M2 x2 f1 1 M1 x1 F

Fig. 3.14

Solution: Step 1: Using nodal analysis at node 1, d 2 x1 (t ) d (x1 − x2 ) F (t) = M1 + f1 2 dt dt Taking Laplace transform on both sides, F (s) = s2 M1 X1 (s) + sf1 (X1 (s) – X2 (s)) Step 2: Using nodal analysis at node 2, d (x2 − x1 ) d 2 x2 (t ) dx (t ) + K 2 x2 (t ) + f 2 2 + f1 0 = M2 2 dt dt dt Taking Laplace transform on both sides, 0 = s2 M2 X2 (s) + K2 X2 (s) + sf2 X2 (s) + sf1 (X2 (s) – X1 (s))

(3.27) (3.28) (3.29) (3.30)

Modeling of Control Systems 55

Thus, equations (3.27) and (3.29) are the system equations in time domain. Equations (3.28) and (3.30) are the system equations in Laplace domain. Step 3: Force-voltage analogous circuit L1

V

L2

R2

R1

1/C 1/ C2

i2 or q2

i1 or q1

Fig. 3.15

Step 4: Force-current analogous circuit 1/R R1 V1 or f1 1/

I

V2 or f2

1/R 1/ R2

C1

1/L 1/ L2

C2

Fig. 3.16

Example 3.2 Obtain system equations and force-voltage and force-current analogies for the translational mechanical system as shown in Fig. 3.17.

f2

K2 K1

f1

2 M2 x2 f3

F 1

M1

x1

Fig. 3.17

Solution: Step 1: Using nodal analysis at node 2, F (t) = M 2

d (x2 − x1 ) d 2 x 2 (t ) dx (t ) + K 2 x2 (t ) + f 2 2 + f 3 2 dt dt dt

(3.31)

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Taking Laplace transform on both sides, F (s) = s2 M2 X2 (s) + K2 X2 (s) + sf2 X2 (s) + sf3 (X2 (s) – X1 (s))

(3.32)

Step 2: Using nodal analysis at node 1, d (x1 − x2 ) d 2 x1 (t ) dx (t ) + K1 x1 (t ) + f1 1 + f 3 2 dt dt dt Taking Laplace transform on both sides,

0 = M1

0 = s2 M1 X1 (s) + K1 X1 (s) + sf1 X1 (s) + sf3 (X1 (s) – X2 (s))

(3.33) (3.34)

Thus, equations (3.31) and (3.33) are the system equations in time domain. Equations (3.32) and (3.34) are system equations in Laplace domain. Step 3: Force-voltage analogous circuit

Fig. 3.18

Step 4: Force-current analogous circuit

Fig. 3.19

Example 3.3 Find the analogous system in terms of torque-voltage and torque-current analogies for the rotational mechanical system as shown in Fig. 3.20. Also, write the system equations.

Fig. 3.20

Modeling of Control Systems 57

Solution: Step 1: At node 1, d (θ1 (t ) − θ 2 (t )) d θ (t ) d 2 θ1 (t ) + f1 1 + K 2 (θ1 (t ) − θ 2 (t )) + f 2 2 dt dt dt Taking Laplace transform on both sides,

T (t) = K1θ1 (t ) + J1

(3.35)

T (s) = K1 θ1 (s) + s 2 J1 θ1 (s) + sf1 θ1 (s) + K 2 (θ1 (s) − θ 2 (s)) + sf 2 (θ1 (s) − θ 2 (s))

(3.36)

Step 2: At node 2, 0 = J2

d (θ 2 (t ) − θ1 (t )) d 2 θ 2 (t ) d θ (t ) + f 3 2 + K3 θ 2 (t ) + K 2 (θ 2 (t ) − θ1 (t )) + f 2 2 dt dt dt

(3.37)

Taking Laplace transform on both sides, 2 0 = s J 2 θ 2 (s) + sf 3 θ 2 (s) + K3 θ 2 (s) + K 2 (θ 2 (s) − θ1 (s)) + sf 2 (θ 2 (s) − θ1 (s))

(3.38)

So, equations (3.35) and (3.37) are system equations in time domain. Equations (3.36) and (3.38) are system equations in Laplace domain. Step 3: Torque-voltage analogous circuit

Fig. 3.21

Step 4: Torque-current analogous circuit

Fig. 3.22

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Automatic Control Engineering

Example 3.4

Draw force-voltage and force-current analogies for Fig. 3.23. K1

f

K2

M1

M2 x1

f1

M3 x2

f2

F x3

f3

Fig. 3.23

Solution: Step 1: Force-voltage analogous circuit R1

I/C1

L1

R

i1 or q1

L2

R2

i2 or q2

L3

R3

i3 or q3

1/C 1/ C2

V

Fig. 3.24

Step 2: Force-current analogous circuit 1/R R V1 or f1 1/ 1/L 1/ L1

1/R 1/ R1

C1

V2 or f2

1/R 1/ R2

C2

1/L 1/ L2

1/R 1/ R3

V3 or f3 C3

I

Fig. 3.25

3.4

THERMAL SYSTEMS

Practically, the temperature of any medium is not constant and so a distributed parameter model should be used for analysis. But for easy analysis, it is assumed that temperature of the medium is constant so that a lumped parameter model can be taken to represent the system. Consider a thermal system as shown in Fig. 3.26. In this system, liquid (water) is heated with the help of heater, in an insulated tank. The heat input rate from the heater is H. The temperature of the liquid is maintained constant with the help of stirrer. The inflow liquid temperature is qi and the outflow liquid temperature is q0. Now, assume that the heat input rate from heater is increased by a small value DH, which will further increase the heat inflow rate by say amount DH1 and heat storage

Modeling of Control Systems 59

Fig. 3.26

in the tank by say amount DH2. This increases the temperature of liquid in tank and hence of outflow liquid by Dq0. (3.39) So, increase in heat inflow rate DH1 = SQDq0 where Q = steady liquid flow rate S = specific heat of liquid

1 Equation (3.39) can be rewritten as DH1 = Dq R 0 1 = thermal resistance of liquid. where R = QS

Again, increase in heat storage in tank, DH2 = SM where M = Mass of liquid in tank Equation (3.41) can be rewritten as DH2 = C

d Δ θ0 dt

(3.40) d Δ θ0 dt

(3.41) (3.42)

where C = MS = thermal capacitance of liquid. Now, heat balance equation becomes, DH = DH1 + DH2 fi

fi

d Δθ0 1 Δθ0 + C R dt Taking Laplace transform on both sides, 1 Dq0 (s) + sC D C 0 (s) DH (s) = R

DH =

Dq0 (s) = DH (s)

1

(3.43)

(3.44) 1 R Equation (3.44) gives the transfer function of the thermal system. Comparing equations (3.21) and (3.43), analogy between electrical and thermal systems is obtained as given in Table 3.3. sC +

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Table 3.3 Electrical System

Thermal System

Voltage (V)

Heat flow rate (H)

Charge (q)

Temperature change (q)

Resistance (R)

Thermal capacitance (C)

Capacitance (C)

Thermal resistance (R)

3.5

HYDRAULIC SYSTEMS

Any hydraulic system can be represented by a set of linear differential equations if the fluid has the following characteristics: (i) Fluid is incompressible (ii) Fluid flow is laminar which means smooth motion of one fluid past another. Consider a liquid level system as shown in Fig. 3.27.

H + DH

Q + DQ

Q0 + DQ0

Fig. 3.27

Let Qi be the input flow rate of liquid and Q0 is the output flow rate of liquid, so that head of the liquid in tank is H. If Qi is changed to Qi + DQi , then corresponding values of Q0 and H are changed to (Q0 + DQ0) and (H + DH) respectively. The fluid pressure is given by equation (3.45). P = rgH (3.45) where P = fluid pressure r = mass density of liquid g = gravity For laminar flow liquid, the pressure-flow rate relations are given by H=

128l m Q pD2 rg

l = length of pipe m = viscosity of liquid D = diameter of pipe Equation (3.46) can be rewritten as H = RQ 128l m where R = Hydraulic resistance = pD2 rg

(3.46)

where

(3.47)

Modeling of Control Systems 61

Also,

DH = RDQ0

DH (3.48) R dH Again, rate of liquid storage in tank = C dt where C = Hydraulic capacitance of the tank Now, system dynamics can be described by liquid flow rate balance equation, as given by equation (3.49). Change in rate of liquid storage in tank = change in rate of inflow liquid – change in rate of outflow liquid

DQ0 =

fi

d DH DH = DQi dt R Taking Laplace transform on both sides, 1 sC D H (s) + DH (s) = DQi (s) R

(3.49)

C

DH (s) = DQi (s)

(3.50)

1

(3.51) 1 sC + R Equation (3.51) gives the transfer function of the hydraulic system. Comparing equations (3.21) and (3.50), analogy between electrical and hydraulic system is obtained as given in Table 3.4. fi

Table 3.4 Electrical System

Hydraulic System

Voltage (V)

Flow rate (Q)

Charge (q)

Head (H)

Resistance (R)

Hydraulic capacitance (C)

Capacitance (C)

Hydraulic resistance (R)

Exercise 1. Develop the system equations for the translational mechanical system shown in Fig. 3.28. Also draw its electrical analogies.

Fig. 3.28

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2. Develop the system equations for the mechanical network shown in Fig. 3.29. 3. Draw the force-voltage and force-current analogies for the system shown in Fig. 3.29.

Fig. 3.29

4. Draw the force-voltage and force-current analogies for the system shown in Fig. 3.30.

Fig. 3.30

5. Write the system equations in time domain and in Laplace domain for the system shown in Fig. 3.30.

Objective Type Questions 1. Mechanical impedance is the ratio of (a) rms velocity to rms force (b) rms force to rms displacement (c) rms force to rms velocity (d) none of these 2. Temperature is analogous to (a) current (b) voltage (c) charge (d) flux 3. In torque-voltage analogy, resistance is analogous to (a) damping coefficient (b) spring (c) angular displacement (d) moment of inertia 4. In torque-voltage analogy, inductance is analogous to (a) damping coefficient (b) spring (c) stiffness (d) moment of inertia

Modeling of Control Systems 63

5. In torque-current analogy, capacitance is analogous to (a) moment of inertia (b) spring (c) angular displacement (d) damping coefficient 6. In force-current analogy, capacitance is analogous to (a) spring (b) moment of inertia (c) angular displacement (d) damping coefficient 7. The concept of analogous systems is applicable to (a) linear systems only (b) non-linear systems only (c) both (a) and (b) (d) none of these 8. In mechanical systems, stiffness force is proportional to (a) velocity (b) displacement (c) acceleration (d) all of these 9. In mechanical systems, spring force is proportional to (a) velocity (b) displacement (c) acceleration (d) all of these 10. The resistance R of a liquid flow is defined as (a) change in head × change in level difference (b) change in flow rate × change in level difference

(c) (d)

change in flow rate change in level difference change in level difference change in flow rate

11. If Q is the heat energy, C is the thermal capacitance and θ is temperature, the electrical analogous system is dv di (b) v = L (a) i = C dt dt df (d) none (c) e = N dt 12. Time rate of change of heat energy is analogous to (a) current (b) voltage (c) charge (d) flux 13. In hydraulic system, the capacitance of water tank is

(a) change in liquid store change in head

(c)

change in flow rate change in head

(b)

change in head change in liquid store

change in head

(d) change in flow rate

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Answers 1. (c) 6. (b) 11. (a)

2. (b) 7. (a) 12. (a)

3. (a) 8. (a) 13. (a)

4. (d) 9. (b)

5. (a) 10. (d)

4

Effects of Feedback on Control Systems

CHAPTER

4.1

INTRODUCTION

As already discussed in Chapter 1, closed loop control systems play an important role in modern engineering as they perform the assigned task more precisely and automatically. Open loop control systems are not automatic. Addition of feedback element to the open loop control system makes it automatic and such automatic control systems are also called feedback control systems. In an open loop control system, the output is not controlled by its input and no mechanism is provided to make the output to follow the input automatically. But in feedback control system, the feedback element continuously senses the output and automatically corrects any deviation of output from input. In this way, output is made to follow input automatically with the help of feedback element. In almost all physical systems, feedback is inherent in nature in control systems. The various parameters of the control system may change due to some conditions. Good control system is that which is insensitive towards these variations. Feedback plays an important role in making the control system insensitive towards these variations. These are explained mathematically in the following sections.

4.2

EFFECT OF FEEDBACK ON OUTPUT DUE TO PARAMETER VARIATIONS IN CONTROL SYSTEM

The parameters in a control system may vary due to different environmental conditions, aging, wear and tear of components, etc. For open loop control system (without feedback), C (s) = G (s) R (s)

(4.1)

Let the system parameter G (s) change to [G (s) + DG (s)], where DG (s) is very small. This corresponds to the change in output from C (s) to [C (s) + DC (s)], so that equation (4.1) becomes, C (s) = [G (s) + DG (s)] R (s)

(4.2)

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From equation (4.1) and equation (4.2), DC (s) = DG (s) R (s) (4.3) Equation (4.3) gives the change in output in non-feedback control system when parameter variations take place. Now, in closed loop control system (with negative feedback), C (s ) =

G (s ) R (s ) 1 + G (s ) R (s )

(4.4)

Again suppose that parameter G (s) changes to [G (s) + DG (s)], where DG (s) is very small. This corresponds to the change in output from C (s) to [C (s) + DC (s)], so that equation (4.4) becomes, C (s) + DC (s) = =

G (s) + DG (s) R (s) 1 + ÈÎG (s) + DG (s)˘˚ H (s) G (s ) R (s ) DG (s) R (s) + 1 + G (s) H (s) + DG (s) H (s) 1 + G (s) H (s) + DG (s) H (s)

Now, DG (s) H (s) >> 1 So, equation (4.20) becomes, fi

C (s ) =

G 2 ( s ) Sd ( s ) Sd ( s ) = G1 (s) G2 (s) H (s) G1 (s) H (s)

(4.21)

From equation (4.21), it is concluded that effect of Sd (s) on C (s) can be reduced if the value of H (s) i.e., feedback is increased. Hence, effect of disturbance present in the forward path is reduced in feedback system as compared to that in non-feedback system.

4.7.2 Disturbance in Feedback Path When the disturbance signal Sd (s) is present in feedback path, the block diagram of the simplest feedback control system is shown in Fig. 4.10. This block diagram is also two input-single output system. To study the effect of disturbance on output signal, R (s) is kept as zero. Thus block diagram in Fig. 4.10 reduces to the block diagram shown in Fig. 4.11. R(s) +

C(s)

G – +

H2 + Sd

Fig. 4.10

H1

Effects of Feedback on Control Systems 75 C(s)

G +

–H2

H1

+ Sd

Fig. 4.11

Here, forward path gain = – G (s) H2 (s) Feedback path gain = H1 (s) C (s ) – G (s ) H 2 (s ) = Sd (s) 1 − ( −G (s) H 2 ( s)) ( H1 (s))

Thus,

C (s ) =

fi

– G ( s ) H 2 ( s ) Sd ( s ) 1 + G (s) H1 (s) H 2 (s)

For further analysis, assume that G (s) H1(s) H2(s) >>>> 1 So, equation (4.22) becomes, – G (s ) H 2 (s ) –1 fi C (s ) = Sd ( s ) = Sd ( s ) G (s) H1 (s) H 2 (s) H1 (s)

(4.22)

(4.23)

From equation (4.23), it is concluded that effect of Sd (s) on C (s) can be reduced if the value of H1 (s) i.e., feedback is increased. Hence, effect of disturbance present in feedback path in the feedback control system reduces as compared to that in nonfeedback control system. Example 4.7 A speed control system for dc motor is shown in Fig. 4.12. The output of the system is speed w (s) and input is voltage V(s). An external disturbance is present in the forward path as Sd(s). Determine sensitivity of the system and where Md is the ratio of w (s) to disturbance signal.

Fig. 4.12

Solution: Step 1:

M d (s ) =

ω (s ) output = input (disturbance) Sd (s)

So, to find Md, the first input signal i.e., V (s) is kept at 0.

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Automatic Control Engineering

Step 2: The new reduced signal flow graph is shown in the Fig. 4.13. Step 3: Forward path gain P1 = G2 = Loop gain L1 = G1 G2 (– Kb) = and

− KK b (s + 2) 2

K ;D =1–0=1 s+2 1

D = 1 – L1 = 1 +

KK b (s + 2) 2

Fig. 4.13

Now, according to Mason’s gain formula, P Δ Md = 1 1 = Δ

K (s + 2) KK b 1+ (s + 2) 2

fi

Md =

K (s + 2) (s + 2) 2 + KK b

(4.24)

Step 4: Now,

SKMd =

∂M d K × ∂K M d

(4.25)

∂M d (s + 2)3 = 2 ∂K ⎡⎣(s + 2) 2 + KK b ⎤⎦

(4.26)

Using equation (4.24),

Putting equations (4.24) and (4.26) in (4.25),

Step 5:

Using equation (4.24),

SKMd =

(s + 2) 2 ⎡⎣(s + 2) 2 + KK b ⎤⎦

SKMb d =

∂M d K b × ∂K b M d

(4.27)

– K 2 (s + 2) ∂M d = ∂K b ⎡(s + 2) 2 + KK ⎤ 2 b⎦ ⎣

(4.28)

(Answer)

Effects of Feedback on Control Systems 77

Putting equations (4.24) and (4.28) in (4.27), – KK b SKMb d = ⎡⎣(s + 2) 2 + KK b ⎤⎦

4.8

(Answer)

CONCLUSIONS

As discussed in previous sections, the advantages of adding feedback in a control system can be summarized as: (i) The effect of parameter variations on output is reduced. (ii) The system becomes more insensitive towards parameter variations. (iii) The dynamic behaviour of the system improves and it becomes faster. (iv) The effect of disturbances on performance of control system reduces. In addition to these advantages, feedback control system has some disadvantages also, which are summarised as: (i) The overall gain of the feedback control system reduces. (ii) The system may become unstable. (iii) The number of components increases and hence complexity. But in all practical applications, the advantages of feedback control system outweigh its disadvantages. Hence, feedback control systems are most commonly used in practical applications.

4.9

REGENERATIVE FEEDBACK

Uptill now, the feedback used in control system is taken as negative feedback which is called degenerative feedback. As already discussed, this negative feedback decreases the overall gain of the control system. So, to increase the overall gain of the control system with the inclusion of feedback, the feedback is fed back with positive sign and such a system is shown in Fig. 4.14.

Fig. 4.14

Here,

C (s ) G (s ) = R (s ) 1 − G (s ) H (s )

(4.29)

Equation (4.29) gives the overall gain of the transfer function for a regenerative feedback control system. This overall gain is more than that of a degenerative feedback control system. As seen from equation (4.29), at some values of G and H, denominator

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may become zero and hence the output of the system may be infinity even for a finite input. This is the condition of instability. Hence, one of the main disadvantages of using regenerative feedback in control system is that the system may become unstable under some values of its parameters.

Exercise 1. Determine the sensitivity of the system shown in Fig. 4.15 for w = 1 rad/s w.r.t. (a) forward path transfer function and (b) feedback transfer function.

Fig. 4.15

2. Consider a feedback control system as shown in Fig. 4.16.

Fig. 4.16

K3

1 qi

E

K1 Ea

–1

1/(s + 1)

Evaluate sensitivity of overall transfer function T (s) to variation in K at w = 2 rad/s. Consider K = 1 3. A servo system is represented by SFG as shown in Fig. 4.17. The variable T is torque and E is error signal. Determine (a) Overall transmission if K1 = 1, K2 = 5 and K3 = 5. (b) The sensitivity of the system w.r.t. at K1 at w = 0 rad/s.

Eb –K2

Fig. 4.17

–5

1/ss T 1/

2

qo

Effects of Feedback on Control Systems 79

4. The block diagram of speed control system is shown in Fig. 4.18. Determine ω (s ) d and SMd where M = SM d K Kb Sd ( s ) . Sd(s) V(s) +

–

1 s+2

+ +

K s+2

w (s)

Kb

Fig. 4.18

Objective Type Questions 1. With the addition of feedback in control system, which of the following improves? (a) Sensitivity (b) Stability (c) Gain (d) None 2. With feedback, which of the following reduces? (a) Gain (b) Stability (c) Sensitivity of system (d) All of these 3. Sensitivity of the control system decreases if feedback sensitivity (a) increases (b) decreases (c) not related (d) remains constant 4. With feedback system (a) transient response remains constant (b) transient response gets magnified (c) transient response decays slowly (d) transient response decays more quickly 5. Regenerative control system is also called (a) no feedback system (b) positive feedback system (c) negative feedback system (d) all of above 6. Degenerative control system is also called (a) no feedback system (b) positive feedback system (c) negative feedback system (d) all of above 7. Time constant is smaller for (a) open loop system (b) closed loop system (c) independent of type of system (d) nothing can be said

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Automatic Control Engineering

8. Construction wise, which system is more complex? (a) Open loop system (b) No feedback system (c) Closed loop system (d) None

Answers 1. (a) 6. (c)

2. (d) 7. (b)

3. (a) 8. (c)

4. (d)

5. (b)

5

Time Response Analysis

CHAPTER

5.1

INTRODUCTION

Every electrical network or system comprises basic network elements like resistor, inductor and capacitor. Out of these three elements, inductor and capacitor are energy storage elements that form the part of the control system. If this control system is disturbed by some perturbation, then the energy state of the system changes from one state to another. So, time response analysis is the study of change in the response or output of the system with respect to time, following a perturbation. The main aim of the control system is to make its output to follow its input, with passage of time. Time response analysis has a great importance in the design part of the control system as these are the inherent characteristics of it. From time response analysis of a control system, its stability can also be studied.

5.2

IMPORTANT TERMS

The time response of a control system has two parts: transient response and steady state response. So, c (t) = cT (t) + css (t) where, c (t) = time response cT (t) = transient response css(t) = steady state response (i) Transient response cT(t): It is that part of time response that goes to zero, as time approaches infinity. The transient response may be oscillatory or exponential. The oscillations may increase or decrease or remain constant with passage of time. cT (t ) = 0 So, tlim →∞ (ii) Steady state response css(t): It is the part of total time response that remains in the system, after transient response vanishes. Figure 5.1 shows the transient and steady state response as part of time response. (iii) Error response e (t): It is the difference between input signal r (t) and output signal c (t) at every instant of time. So, e (t) = r (t) – c (t).

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Automatic Control Engineering

Fig. 5.1

(iv) Steady state error ess: It is the error between input signal r (t) and output signal c (t), as time approaches infinity or under steady state. Mathematically, ess = lim e (t ) t→∞

5.3

STANDARD TEST INPUT SIGNALS FOR TRANSIENT ANALYSIS

The input signals that act as reference signals in the control systems are known as standard test signals. The physically realisable signals that are used to study time response analysis are as follows:

5.3.1

Step Signal

It is the signal whose value changes from one level to another level in negligible time. Usually, the initial level is zero level and final level is some constant level. If that constant level is one, then it is called unit step signal. Mathematically, r (t) = A; t ≥ 0 = 0; t < 0 In Laplace transform, R (s) = A/s If A = 1, then r (t) = u (t), unit step signal and U (s) = 1/s It is graphically shown in Fig. 5.2(a).

5.3.2

Ramp Signal

It is the signal that varies linearly with time. Mathematically, r (t) = At; t ≥ 0 = 0; t < 0 In Laplace transform, R (s) = A/s2 If A = 1, then r (t) is called unit ramp signal and R (s) = 1/s2 It is graphically shown in Fig. 5.2(b).

Time Response Analysis 83

5.3.3

Parabolic Signal

It is the signal that varies quadratically with time. Mathematically, r (t) = At2/2; t ≥ 0 = 0; t < 0 In Laplace transform, R (s) = A/s3 If A = 1, then r(t) is called unit parabolic signal and R (s) = 1/s3 It is graphically shown in Fig. 5.2(c).

Fig. 5.2

5.3.4

Impulse Signal

It is a special type of pulse signal for which duration tends to zero and amplitude tends to infinity. Mathematically, d (t) = 0; t ≠ 0 = •; t = 0 So, impulse signal has zero value everywhere except at t = 0, where magnitude is infinite. Also, impulse function is derivative of unit step function, i.e., d (t) = 0; t ≠ 0 du (t ) ; t=0 dt It is graphically shown in Fig. 5.2(d). In Laplace transform, d (s) = s (1/s) = 1

=

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Automatic Control Engineering

On the basis of all discussed test signals, a relational diagram can be derived, which is shown in Fig. 5.3.

Fig. 5.3

5.4

TIME RESPONSE OF FIRST ORDER SYSTEM

Consider a degenerative system as shown in Fig. 5.4(a). Its overall transfer function is shown in Fig. 5.4(b). As the highest power of s in the denominator of transfer function is one, so it is an example of first order system.

Fig. 5.4

Here, transfer function is C (s ) 1 = R (s) 1 + Ts

(5.1)

Now, different test signals will be applied to this system and behaviour of output signal and error signal will be studied.

5.4.1 Time Response to the Unit Step Input From equation (5.1), C (s ) 1 = R (s) 1 + Ts

Applying unit step input i.e., U (s) = 1/s,

C (s) =

1 1 T = – s ( 1 + Ts ) s 1 + Ts

Taking Laplace inverse on both sides, (5.2) c (t) = 1 – e – t/ T Equation (5.2) is plotted in Fig. 5.5(a). It is exponential in nature as is also seen from equation (5.2).

Time Response Analysis 85

Fig. 5.5

Putting t = T in equation (5.2), c (t) = 0.632 or 63.2%. This T is known as ‘time constant’ of the system and is defined as the time required by the output signal c (t) to reach 63.2% of its final value or steady state value (here 1). It is an indicative of how fast the system response reaches its final value. So, it quantitatively defines the speed of the response. Smaller the time constant T, faster is the response of the system and better is the transient response. As T increases, the system response becomes slow and sluggish. The output response corresponding to two time constants T1 and T2 is compared in Fig. 5.5(b). Now, error signal e (t) = r (t) – c (t) (5.3) = 1 – (1 – e – t/T ) = e – t /T This error signal in equation (5.3) is also plotted in Fig. 5.5(a). Steady state error is ess = lim e (t ) = 0 t→∞

Thus, under steady state, c (t) becomes equal to r (t) and steady state error becomes zero.

5.4.2 Time Response to the Unit Ramp Input From equation (5.1), C (s ) 1 = R (s) 1 + Ts

Applying unit ramp input i.e., R (s) = 1/s2, 1 1 T T2 = 2 – + C (s) = 2 s 1 + Ts s ( 1 + Ts ) s Taking Laplace inverse on both sides, c (t) = t – T + e – t/T fi c (t) = t – T (1 – e – t/T ) Equation (5.4) is plotted in Fig. 5.6. Here, error signal e (t) = r (t) – c (t) = T (1 – e – t/ T) = e – t/T

(5.4)

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Automatic Control Engineering

Steady state error is ess = lim e (t ) = T

(5.5)

t→∞

Equation (5.5) is also plotted in Fig. 5.6. It is clear that c (t) will follow r (t) in steady state but with constant error and that is equal to the time constant (T) of the system. Also, smaller the time constant T, smaller is the steady state error and better is the system.

Fig. 5.6

Fig. 5.7

5.4.3 Time Response to the Unit Parabolic Input From equation (5.1), C (s ) 1 = R (s) 1 + Ts

Applying unit parabolic input i.e., R (s) = 1/s3, 1 1 T T2 T3 = 3 – 2 + – C (s) = 3 s 1 + Ts s ( 1 + Ts ) s s Taking Laplace inverse on both sides, c (t) = t 2/2 – Tt + T 2 + T 2 e– t /T c (t) = t2/2 – Tt + T 2 (1 – e – t/T )

fi

(5.6)

Equation (5.6) is plotted in Fig. 5.7. Here, error signal e (t) = r (t) – c (t) = Tt – T 2 (1 – e –t / T) Steady state error is ess = lim e (t ) = • t→∞

(5.7)

Equation (5.7) is also shown in Fig. 5.7. c (t) will follow r (t) in steady state but the error between these two goes on increasing and reaches infinity.

5.4.4 Time Response to the Unit Impulse Input From equation (5.1), C (s ) 1 = R (s) 1 + Ts

Time Response Analysis 87

Applying unit impulse input i.e., R (s) = 1, 1 C (s) = (1 + Ts ) Taking Laplace inverse on both sides, 1 – t /T e c (t) = T Equation (5.8) is plotted in Fig. 5.8. Now error signal d 1 u(t ) – e– t /T e (t) = dt T

(5.8)

Steady state error is ess = lim e (t ) = 0

(5.9)

t→∞

Equation (5.9) is also shown in Fig. 5.8.

Fig. 5.8

So, c (t) follows r (t) [zero under steady state], under steady state with steady state error equal to zero. Example 5.1 Consider a negative feedback system shown in Fig. 5.9. Find the expression for output and error when the system is subjected to unit step signal. + r(t)

–

e(t)

1 1 + 2s

c(t)

Fig. 5.9

Solution: Step 1: The overall transfer function is

C (s) G (s) = R (s ) 1+ G (s ) H (s ) =

1 0.5 = 2s + 2 s + 1

(5.10)

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Step 2: Comparing equation (5.10) with equation (5.2), c (t) = 0.5 (1 – e – t/1) = 0.5 – 0.5 e – t

(5.11)

Step 3: Comparing equation (5.10) with equation (5.3), e (t) = 1 – 0.5 + 0.5 e –t/1 = 0.5 + 0.5 e – t

(5.12)

Equations (5.1) and (5.12) are plotted in Fig. 5.10.

Fig. 5.10

Example 5.2 Determine the response to unit step of speed control system shown in Fig. 5.11. Also, find the steady state error.

Fig. 5.11

Solution: Step 1: The overall transfer function for the system is:

5 C (s ) 1 + 2s 5 0.833 = = = R (s ) 1 + 5 2 s + 6 0.33 s +1 1 + 2s Step 2: Comparing equation (5.13) with equation (5.2), c (t) = 0.833 (1 – e – t/0.33)

(5.13)

(Answer)

Step 3: Comparing equation (5.13) with equation (5.3), e (t) = 1 – 0.833 + 0.833 e– t/0.33 = 0.167 + 0.833 e –t/00.33 Step 4:

ess = lim e (t ) e (t) = 0.167 t→∞

(Answer)

Time Response Analysis 89

5.5

TIME RESPONSE OF SECOND ORDER SYSTEM

Consider a degenerative system as shown in Fig. 5.12(a). Its overall transfer function is shown in Fig. 5.12(b). As highest power of s in the denominator of transfer function comes out to be 2, it is an example of second order system.

Fig. 5.12

Here,

2 n

C (s ) ω = 2 R (s) s + 2ω n δs + ω n2

(5.14)

Now, two types of test signals i.e., unit step and unit ramp are applied to it, to study time response analysis.

5.5.1 Time Response to Unit Step Input From equation (5.14),

C (s ) ω n2 = 2 R (s) s + 2ω n δs + ω n2

Applying unit step input i.e., U (s) = 1/s, − s − 2 δω n ω n2 1 = + 2 C (s) = 2 2 s s + 2 δω n s + ω n2 s s + 2 δω n s + ω n

(

)

s + 2 δω n 1 − 2 s s + 2 δω n s + ω n2 − ω n2 δ 2 + ω n2 δ 2 s + 2 δω n 1 = − s (s + δω n )2 + ω n2 ( 1 − δ 2 ) =

Now, put

ω n2 ( 1 − δ 2 ) = ω d2 ωd = ωn

fi Thus,

(1 − δ ) 2

s + 2 δω n 1 C (s ) = – s (s + δω n )2 + ω d2

where wn is undamped natural frequency, d is damping ratio or damping factor, wd is damped frequency of oscillations. Again, 1 s + δω n + δω n C (s) = – s (s + δω n )2 + ω d2

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⎡ ⎤ δω n ω d s + δω n 1 ⎢ ⎥ + = –⎢ 2 2 s s + δω n ) + ω d2 ω d ⎡(s + δω n ) + ω d2 ⎤ ⎥ ( ⎣ ⎦ ⎦⎥ ⎣⎢

Taking Laplace inverse transform on both sides, ⎛ ⎞ δ – δω t e – δωn t sin ω d t⎟ c (t) = 1 – ⎜ e n cos ω d t + ⎜⎝ ⎟⎠ 1 – δ2 = 1–

e – δωn t 1– δ

2

( 1– δ

2

cos ω d t + δ sin ω d t

)

Now, consider a right angled triangle as shown in Fig. 5.13 1

1 – d2

q d

Fig. 5.13

Here, sin q =

1 – δ 2 and cos θ = δ

Therefore, c (t) = 1 – = 1–

fi

c (t) = 1 –

e – δωn t 1 – δ2 e – δωn t 1 – δ2

(sin θ cos ω t + cos θ sin ω t ) d

d

sin (ω d t + θ )

⎛ 1 – δ2 ⎞ sin ⎜ ω n 1 – δ 2 t + tan – 1 ⎟ δ ⎠ 1 – δ2 ⎝

e – δωn t

Equation (5.15) is plotted in Fig. 5.14.

Fig. 5.14

(5.15)

Time Response Analysis 91

Here,

e (t) = u (t) – c (t) ⎛ 1 – δ2 ⎞ sin ⎜ ω n 1 – δ 2 t + tan – 1 ⎟ δ ⎠ 1 – δ2 ⎝

e – δωnt

e (t) =

fi

Steady state error is ess = lim e (t ) = 0

(5.16)

t→∞

Equation (5.16) is also plotted in Fig. 5.14. So, c (t) will follow u (t) and becomes equal to c (t) in steady state thereby making steady state error equal to zero. Normally, all the control systems are designed for damping factor (d) less than 1. So, equation (5.15) gives the time response to unit step input for second order underdamped system. For critically damped system i.e., for d = 1, ω n2 ω n2 1 = C (s) = . 2 2 s s + 2δω n s + ω n2 s (s + ω n )

(

=

)

ωn 1 1 – – s s + ω n ( s + ω n )2

Taking Laplace inverse transform on both sides, c (t) = 1 – e – wnt – t wn e – wnt (5.17) fi c (t) = 1 – e – wnt (1 + wnt) Equation (5.17) gives the time response to unit step input for second order critically damped system. For overdamped system i.e., for d > 1, w n2 w n2 = C (t) = 2 s s 2 + 2 d w n s + w n2 s ⎡( s + d w n ) – w d2 ⎤ ⎣ ⎦ 1 1 1 – = + 2 2 2 s 2 – d +1 ⎡d + – d +1⎤ (s + dw + w ) 2 – d +1 ⎡d – – d 2 +1⎤ ( s + dw – w ) n d n d ⎣ ⎦ ⎣ ⎦

(

1

=s +

1 1 – 2 2 2 2 ⎡ ⎤ ⎡ 2 – d +1 d + – d +1 s + dw n + w n – d +1 2 – d +1 d – – d +1⎤ s + dw n – w n – d 2 +1 ⎣ ⎦ ⎣ ⎦

= 1+ s

)

(

2

)

(

)

1 1 – 2⎤⎡ 2 ⎤ 2 ⎡ 2⎤ ⎡ ⎡ 2 1 – d d + 1 – d ⎢s + w n d + 1 – d ⎥ 2 1 – d d – 1 – d ⎢s + w n d – 1 – d 2 ⎤⎥ ⎣ ⎦⎣ ⎣ ⎦⎣ ⎦ ⎦

(

2

)

(

)

Taking Laplace inverse transform on both sides, c (t) = 1 +

e

(

)

– δ + 1 − δ 2 ωn t

2 1 – δ 2 ⎡δ + 1 − δ 2 ⎤ ⎣ ⎦

–

e

(

)

– δ – 1 – δ 2 ωn t

2 1 – δ 2 ⎡δ − 1 − δ 2 ⎤ ⎣ ⎦

(5.18)

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Automatic Control Engineering

According to equation (5.18), there are two time constants in the system, 1 1 and T2 = T1 = ω n ⎡δ + 1 − δ 2 ⎤ ω n ⎡δ − 1 − δ 2 ⎤ ⎣ ⎦ ⎣ ⎦ T1 corresponds to first exponential term and T2 corresponds to second exponential term in equation (5.18). Now, T1 is smaller than T2 i.e., the first exponential term decays much faster than other exponential term. So, T1 can be neglected as compared to T2. Equation (5.18) reduces to c (t) = 1 − with time constant T=

e

(

)

– δ − 1 − δ 2 ωn t

2 1 – δ 2 ⎡δ − 1 − δ 2 ⎤ ⎣ ⎦

(5.19)

1 ω n ⎡δ − 1 − δ 2 ⎤ ⎣ ⎦

So, equation (5.19) gives the time response to unit step input for second order overdamped system. Due to large time constant, the response becomes sluggish. The response is shown in Fig. 5.15.

Fig. 5.15

For undamped system, d = 0 Equation (5.15) reduces to c (t) = 1 – sin (wnt + tan–1 •) = 1 – sin (wnt + p/2) fi

c (t) = 1 – cos wnt

(5.20)

Equation (5.20) gives the time response to unit step input for second order undamped system. Graphically, it is shown in Fig. 5.16.

5.5.2 Time Response to Unit Ramp Input From equation (5.14), transfer function of second order system is

C (s ) ω n2 = 2 R (s ) s + 2 δω n s + ω n2

Time Response Analysis 93

Fig. 5.16

Applying unit ramp input i.e., R (s) = 1/s2, ω n2 C (s) = 2 2 s s + 2 δω n s + ω n2

(

)

During the discussion of time response of second order system to unit step signal, it is seen that (s2 + 2dwn s + wn2 ) can be rewritten as [(s + dwn)2 + wd2]. So, here 2δ s + (4δ 2 – 1) ω ω 1 2δ = 2 – + n C (s) = 2 2 2 ω s s s + δω n ) 2 + ω d2 ( ⎡ ⎤ s ⎣(s + δω n ) + ω d ⎦ n 2 n

2δ (s + δω n ) + (2δ 2 – 1) ωn 1 2δ + = 2 – s ωn s (s + δω n ) 2 + ω d2 (s + δω n ) ωd 1 2δ 2δ (2δ 2 – 1) – + + = 2 s ω n s ω n [(s + δω n ) 2 + ω d2 ] ωd [(s + δω n ) 2 + ω d2 ]

Taking Laplace inverse transform on both sides, c (t) = t –

( 2δ 2 – 1) e – δωnt sin ω t 2δ 2δ – δωn t + cos ω d t + e d ωn ωn ωd

fi

c (t) = t –

2δ 2 – 1 – δω t 2δ 2δ – δωn t e e n sin ω n 1 – δ 2 t + cos ω n 1 – δ 2 t + ωn ωn ωd

fi

c (t) = t –

2δ e – δωn t ⎡ + 2δ 1 – δ 2 cos ω n 1 – δ 2 t + 2δ 2 – 1 sin ω n 1 - δ 2 t ⎤ 2 ⎣ ⎦ ωn ωn 1 – δ

(

)

(

Now, consider a right angled triangle, as shown in Fig. 5.17.

2d

1–d

1

2

q 2

2d – 1

Fig. 5.17

)

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Automatic Control Engineering

Here, sin q = 2 d√1 – d2 and cos q = 2 d2 – 1 2δ e – δωn t ⎡ + sin θ cos ω n 1 – δ 2 t + cos θ sin ω n 1 – δ 2 t ⎤ c (t) = t – ⎦ ωn ωn 1 – δ 2 ⎣ So, where Thus,

= t–

(

2δ e – δωnt + sin ω n 1 – δ 2 t + θ ωn ωn 1 – δ 2 −1 q = tan

c (t) = t –

2 δ 1 – δ2 2 δ2 − 1

)

= 2 tan −1

1 – δ2 δ

⎛ 1 – δ2 ⎞ 2δ e sin ⎜ ω n 1 – δ 2 t + 2 tan −1 + ⎟ ⎜⎝ ωn ωn 1 – δ 2 δ ⎟⎠ – δω n t

(5.21)

Fig. 5.18

Equation (5.21) gives the time response of second order system to unit ramp input. It is plotted in Fig. 5.18. Also, ⎛ 1 – δ2 ⎞ e – δωnt 2δ sin ⎜ ω n 1 – δ 2 t + 2 tan −1 + ⎟ e (t) = r (t) – c (t) = ⎜⎝ ωn ωn 1 – δ 2 δ ⎟⎠ 2δ (constant) (5.22) ωn Equation (5.22) gives the steady state error which comes out to be a constant value and is also plotted in Fig. 5.18. Example 5.3 Obtain the unit step response of a unity feedback system whose open loop 10 . Also, find damping factor and natural frequency. transfer function is G (s) = s (s + 10) Discuss the nature of the system.

and

e (t ) = ess = lim t →∞

Solution: Step 1: Overall transfer function is

C (s) G (s) = R (s) 1 + G (s ) H (s )

Time Response Analysis 95

10 s ( s +10) 10 = 2 = 10 s +10 s +10 1+ s ( s +10)

(5.23)

Step 2: Comapiring equation (5.23) with equation (5.14), (Answer) wn = √10 = 3.162 rad/s Step 3: Also, 2dwn = 10 fi d = 1.58 (Answer) Step 4: As d > 1, the system is overdamped in nature. Example 5.4 Determine the values of G and K such that the system has d of 0.7 and wn of 4 rad/s for the system shown in Fig. 5.19.

Fig. 5.19

Solution: Step 1: The overall transfer function is

G s ( s + 2) C (s ) = R (s) 1 + 1 + Ks ( ) s (sG+ 2) G G = 2 = 2 s + 2 s + G + GKs s + s ( 2 + GK ) + G

(5.24)

Step 2: Comparing equation (5.24) with equation (5.14), wn = √G = 4 fi G = 16 (Answer) Step 3: Also, 2dwn = 2 + GK fi 2 × 0.7 × 4 = 2 + 16 (K) fi K = 0.225 (Answer) Example 5.5 Block diagram of a space vehicle control system is shown in Fig. 5.20. Considering time constant of the controller as 3 sec and ratio of torque to inertia i.e., Tr /J as 2/9 rad2/sec2, find d of the system.

Fig. 5.20

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Solution: Step 1: Characteristic equation is 1 + G (s) H (s) = 0 fi

⎡Tr (Ts + 1)⎤⎦ 1+ ⎣ =0 Js 2

fi

⎡Tr (3 s + 1)⎤⎦ 1+ ⎣ =0 Js 2

fi

Js 2 + Tr (3 s + 1) = 0

Tr (3 s + 1) = 0 J 2 s 2 + (3 s + 1) = 0 fi 9 2 2 2 fi s + s+ =0 3 9 Step 2: Compairing equation (5.25) with s2 + 2dwn s + wn2 = 0; s2 +

fi

(5.25)

2dwn = 2/3 and wn = √(2/9) = 0.471 rad/s d = 0.707

fi

5.6

(Answer)

TIME RESPONSE SPECIFICATIONS OF SECOND ORDER SYSTEM

Second and higher order control systems generally have a pair of complex conjugate poles with damping factor less than one and these poles dominate over all other poles. Thus, the time response of second and higher order control systems generally consists of damped oscillations when subjected to unit step input, as shown in Fig. 5.21. It is clear from the figure that the step response has a number of undershoots and overshoots that decay exponentially and finally the response reaches to steady state. Such step response is characterised by various time response specifications which are related to each other. Quantitatively, these are understood as: how fast the response practically reaches the final value, how much oscillatory the response is, how much time does the response take to reach the input at first, etc. These are graphically shown in Fig. 5.21. Mathematically, these are defined as follows: (i) Delay time (Td): The time required by the response to reach 50% of the final value, in first attempt is known as delay time. (ii) Rise time (Tr): The time required by the step response to rise from 0% to 100% of final value is called rise time for underdamped system. It is shown in Fig. 5.21. The time required by the response to rise from 10% to 90% of final value is called rise time for overdamped system. (iii) Peak time (Tp): The time required by the response to reach first peak overshoot is called peak time.

Time Response Analysis 97

Fig. 5.21

(iv) Settling time (Ts): The time required by the response to reach and stay within specified tolerance band (normally ±2% or ±5%) of its final value is called settling time. Tolerance band generally depends upon the value of damping factor of the system. (v) Maximum overshoot (Mp): The maximum difference between the first peak of step response and the steady state output is called maximum overshoot. It is also c (Tp ) − c (∞ ) × 100 defined as % Mp = c (∞ ) For second order system subjected to unit step input, c (•) = 1 fi % Mp = (c (Tp) – 1) × 100% (vi) Steady state error (ess): The difference between step response and step input when time t tends to infinity or in steady state is called steady state error.

ess = lim ⎡⎣r (t ) − c (t )⎤⎦ t→∞ It becomes clear that by defining all these specifications, the behaviour of the step response of control system can be understood, which is very important for designing aspect of the control systems. So, it becomes more important to define some mathematical relationships for these specifications in order to estimate their values. It should also be noted that these specifications are mutually dependent on each other. The mathematical expression for step response for second order system for underdamped system to unit step input is given by equation (5.15). (i) Rise time (Tr): It is obtained when c (t) reaches unity at the first attempt. So, putting c (t) = 1 and replacing t by Tr , equation (5.15) becomes,

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Automatic Control Engineering

1 = 1–

⎛ 1 – δ2 ⎞ sin ⎜ ω n 1 – δ 2 Tr + tan – 1 ⎟ ⎜⎝ δ ⎟⎠ 1 – δ2

e – δωn Tr

fi

⎛ 1 – δ2 ⎞ sin ⎜ ω n 1 – δ 2 Tr + tan –1 ⎟ =0 ⎜⎝ δ ⎟⎠

fi

ω n 1 – δ 2 Tr + tan – 1

1 – δ2 δ

Put n = 1,

Tr =

π − tan –1 ωn

= nπ

1 – δ2

δ 1 – δ2

(5.26)

seconds

Equation (5.26) gives the mathematical expression for rise time. (ii) Peak time (Tp): It is the time at which c (t) attains maximum value. So, applying the condition of maxima on equation (5.15), i.e., differentiating c (t) w.r.t. t and putting it equal to zero and replacing t by Tp, equation (5.15) becomes 0=0–

fi

e

⎛ ⎛ 1 – δ2 ⎞ 1 – δ 2 ⎞ e – δωn Tp –1 2 2 cos ⎜ ω n 1 – δ 2 Tp + tan – 1 (δω n ) ⎟ ω n 1 – δ − sin ⎜ ω n 1 – δ Tp + tan ⎟ ⎜⎝ δ ⎟⎠ δ ⎟⎠ 1 – δ 2 1– δ ⎝⎜ – δω n Tp

2

⎛ ⎛ 1 – δ2 ⎞ 1 – δ2 ⎞ –1 –1 2 2 2 δ sin ⎜ ω n 1 – δ Tp + tan ⎟ − 1 – δ cos ⎜ ω n 1 – δ Tp + tan ⎟ =0 ⎜⎝ ⎜⎝ δ ⎟⎠ δ ⎟⎠ Now, consider a right angled triangle as shown in Fig. 5.22. 1–d

1

2

q d

Fig. 5.22

Here,

sin q =

1 – δ and cos θ = δ and tan θ = 2

Therefore, equation (5.27) reduces to fi fi fi

(

)

(

1 – δ2 δ

)

cos θ sin ω n 1 – δ 2 Tp + θ − sin θ cos ω n 1 – δ 2 Tp + θ = 0

( sin ( ω

)

sin ω n 1 – δ 2 Tp + θ − θ = 0 n

)

1 - δ 2 Tp = 0

(5.27)

Time Response Analysis 99

fi

ω n 1 – δ 2 Tp = nπ

Put n = 1,

Tp =

π ωn 1 – δ 2

seconds

(5.28)

Equation (5.28) gives the mathematical expression for peak time. (iii) Settling time (Ts ): For an underdamped second order system, the step response is given by equation (5.15). This equation has two components: exponentially decaying component i.e., e

−δω n t 1 − δ2

and sinusoidal oscillatory component

⎡ 1 − δ2 ⎤ –1 2 i.e., sin ⎢ω n 1 − δ t + tan ⎥ . This step response oscillates between a δ ⎥ ⎢⎣ ⎦ pair of envelops before settling down at steady state. The time constant of these envelops is same as the time constant of exponential term of equation (5.15) i.e., 1/dwn. The time required by the step response to reach and stay within these tolerance bands is called settling time. It should be noted that the oscillations do not go outside this tolerance band after settling time is reached. Settling time is defined as Ts = 4 × time constant = 4 × 1/dwn = 4/s (for ± 2% tolerance band) or 3 × time constant = 3 × 1/dwn = 3/s (for ± 5% tolerance band) where s is attenuation in the system. (iv) Maximum overshoot (Mp): For second order underdamped system subjected to unit step input, Mp is given by Mp = c ( Tp ) – 1 e

= 1–

= –

e

Now,

Tp =

⎛ 1 – δ2 ⎞ sin ⎜ ω n 1 – δ 2 Tp + tan –1 ⎟ −1 ⎜⎝ δ ⎟⎠ 1 – δ2 – δω n Tp

⎛ 1 – δ2 ⎞ sin ⎜ ω n 1 – δ 2 Tp + tan –1 ⎟ ⎜⎝ δ ⎟⎠ 1 - δ2

– δω n Tp

π ωn 1 – δ 2 – δω n π

fi

Mp = –

ωn 1 – δ 2

e

1 – δ2 – δπ

= –

e

1 – δ2

1 – δ2

⎛ 1 – δ2 ⎞ π 2 –1 sin ⎜ ω n 1 – δ + tan ⎟ ⎜⎝ δ ⎟⎠ ωn 1 – δ 2

⎛ 1 – δ2 ⎞ sin ⎜ π + tan –1 ⎟ ⎜⎝ δ ⎟⎠

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Automatic Control Engineering – δπ

= +

1 – δ2

e

1 – δ2

From Fig. 5.22,

⎛ 1 – δ2 ⎞ –1 sin ⎜ tan ⎟ ⎜⎝ δ ⎟⎠

⎛ 1 – δ2 ⎞ 1 – δ 2 and tan –1 ⎜ ⎟ =θ ⎜⎝ δ ⎟⎠

sin q = Thus,

– δπ

Mp =

1 – δ2

e

1– δ

2

1 – δ2

– δπ 1 – δ2

fi Mp = e (5.29) Equation (5.29) gives the mathematical expression for maximum overshoot. Example 5.6 Find the time response specifications of the closed loop transfer function given by

C (s ) 10 = 2 R (s) s + 15 s + 100

Solution: Step 1: Comparing the given transfer function with equation (5.14), wn = √100 = 10 rad/s and

2dwn = 15 fi d = 0.75

Step 2: Rise time

Tr =

Step 3: Peak time

Tp =

Step 4:

Ts =

π − tan −1 1 − δ 2 ωn 1 − δ 2 π ωn 1 − δ 2

= 0.343 s

= 0.476 s

4 = 0.533 s (For ±2% tolerance band) δω n

(Answer) (Answer) (Answer)

− δπ

Step 5:

Maximum overshoot (Mp) = e

1 − δ2

= 0.028 = 2.8%

(Answer)

Example 5.7 A unity feedback system is characterised by an open loop transfer function K . Determine gain K so that system will have d of 0.5. For this value of K, s (s + 10) determine peak overshoot, time to peak overshoot and settling time for unit step input.

G (s) =

Time Response Analysis 101

Solution: Step 1: Overall transfer function is

C (s ) K = 2 R (s) s + 10 s + K

Characteristic equation is s2 + 10 s + K = 0. Comparing it with equation (5.14), wn = √K rad/s and

2dwn = 10 fi 2 × 0.5 × √K = 10 fi K = 100

(Answer)

− δπ 1 − δ2

= 0.163 = 16.3% Step 2: Maximum overshoot (Mp) = e π Step 3: Peak time Tp = = 0.362 s ωn 1 − δ 2 4 0.8 s (For ± 2% tolerance band) Step 4: Ts = δω n

(Answer) (Answer) (Answer)

Example 5.8 A closed loop system is shown in Fig. 5.23. Find A1 and A2 so that Mp = 25% and Tp = 4 s, when system is subjected to unit step input.

Fig. 5.23

Solution: Step 1: Overall transfer function is

C (s ) A1 = R (s) s 2 + A1 A2 s + A1

Comparing it with equation (5.14), wn = √A1 rad/s and 2dwn = A1A2 fi d = (A2√A1) / 2

(Answer)

− δπ

Step 2: Maximum overshoot (Mp) = e On solving,

A2 A1 4 − A1 A22

Step 3: Peak time Tp = On solving,

= 0.44 π ωn 1 − δ 2

1 A1

4 − A1 A22

1 − δ2

= 0.25 (5.30)

= 4s

= 0.637

(5.31)

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Step 4: Solving equations (5.30) and (5.31), A1 = 0.739 and A2 = 0.9405

5.7

(Answer)

DESIGN CONSIDERATIONS: PERFORMANCE INDICES

In previous section, various specifications of the control systems are defined that describe the overall performance of the system. The control system is made optimum if all these specifications (Td, Tr, Tp, Ts, Mp and ess) are satisfied precisely and simultaneously. This task is a hit and trial procedure and becomes very troublesome for a design engineer. This task however becomes logical, easier and straightforward if it is possible to establish a single performance index for a system that judges its quality. So, performance index is an important characteristic of the system that measures system’s performance, quality and goodness. The various desirable characteristics of performance index are: (i) Selectivity: It must clearly conclude that the system is optimum or non-optimum. (ii) Sensitivity: It should be very sensitive to the parameter that defines it. (iii) Easy to compute: It should be easily computable and must not be too complex. The general form of performance index is given by equation (5.32). ∞

J=

∫ f (e) dt

(5.32)

0

where J is performance index, f (e) is error function. Various performance indices for LTI systems are: (i) Integral square error (ISE): According to this criterion, J is given by: ∞

J=

∫e

2

(t ) dt

0

where e (t) is the error function of the system. It has many advantages like it is easily computable and it includes both the negative and positive values of the error. But it has a poor sensitivity to parameter variations. (ii) Integral absolute error (IAE): According to this criterion, J is given by: ∞

J=

∫ e (t) dt 0

It is the most simplified and easily applied index. (iii) Integral time absolute error (ITAE): According to this criterion, J is given by: ∞

J=

∫ t e (t) dt 0

It reduces the weighting of larger initial errors and penalizes the small errors occurring later more heavilty that makes this index very selective in nature.

Time Response Analysis 103

(iv) Integral time-square error (ITSE): According to this criterion, J is given by: ∞

J=

∫ te

2

(t ) dt

0

In comparison to other indices, this is difficult to compute but it is somewhat less sensitive. It is already discussed that the performance index of the system depends upon some system parameter value. The performance index which is most insensitive to this system parameter variation is the most effective performance index. By considering that system parameter as damping actor (d), comparison can be drawn between all the four already defined performance indices. Different performance indices for various damping factors for second order system are shown in Fig. 5.24. The characteristics of all the performance indices are summarised in Table 5.1.

Fig. 5.24

Table 5.1 Type of performance index

Selectivity

Minimum value of δ

ISE

Poor

0.5

IAE

Better than ISE

0.7

ITAE

Best

0.707

ITSE

Less than IAE

0.6

5.8

CONCEPT OF STEADY STATE ERROR

Steady state error (ess) is one of the most important specifications of the control system that is directly related to its performance, quality and accuracy. It is defined as the difference between the actual output and desired output when the system has reached

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its steady state. Undoubtedly, the system that has lesser value of ess is more accurate, good in quality and gives better performance. There are many sources of errors in a control system like nature of inputs, types of system, non-linear components such as static friction, backlash, etc., which are further increased by drifts, aging or deterioration. Consider a negative feedback system as shown in Fig. 5.25.

Fig. 5.25

Here, error signal is E (s), which can be written as: E (s) = R (s) – B (s) = R (s) – C (s) H (s) = R (s) – E (s) G (s) H (s) fi E (s) [1 + G (s) H (s)] = R (s) 1 R (s ) fi E (s) = 1 + G (s ) H (s ) Now, steady state error ess = lim e (t ) = lim sE (s) (Using Final Value Theorem) t→∞

s→0

sR (s) ess = lim s → 0 1 + G (s ) H (s )

Thus,

(5.33)

From equation (5.33), it is clear that the steady state error depends upon (i) Type of input R (s) (ii) Product of G (s) H (s) i.e., type of the system.

5.9

EFFECT OF CHANGING INPUTS ON STEADY STATE ERROR

Different types of inputs are applied to equation (5.33) and expressions of ess are derived.

5.9.1 Unit Step Input r (t) = u (t) = 1 fi (s) = 1/s So, equation (5.33) becomes, s (1 / s) 1 = ess = lim s → 0 1 + G (s ) H (s ) 1 + lim G (s) H (s) s→0

For a particular system, the term lim G (s ) H (s ) is constant and is called position s→0

error constant (Kp). Thus,

ess =

1 1 + Kp

(5.34)

Time Response Analysis 105

where

Kp = lim G (s ) H (s ) s→0

So, step response of second order underdamped system is shown in Fig. 5.26.

Fig. 5.26

5.9.2 Unit Ramp Input r (t) = t fi R (s) = 1/s2. So, equation (5.33) becomes, s (1 / s 2 ) 1 1 1 = lim = = ess = lim s → 0 1 + G (s) H (s ) s → 0 s (1 + G (s ) H (s ) ) lim [s + sG (s) H (s)] lim sG (s) H (s) s→0

s→0

For a particular system, the term lim sG (s ) H (s ) is constant and is called velocity s→0 error constant (Kv). 1 (5.35) Thus, ess = Ka where Kv = lim s 2 G (s ) H (s ) s→0

So, ramp response of second order underdamped system is shown in Fig. 5.27.

Fig. 5.27

5.9.3 Unit Parabolic Input r (t) = t 2/2 fi R (s) = 1/s3. So, equation (5.33) becomes, s (1 / s3 ) 1 1 1 = lim 2 = = ess = lim 2 2 2 s → 0 1 + G (s ) H (s ) s → 0 s (1 + G (s ) H (s ) ) lim [s + s G (s) H (s)] lim s G (s) H (s) s→0

s→0

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Automatic Control Engineering

For a particular system, the term lim s 2 G (s ) H (s ) is constant and is called acceleration s→0 error constant (Ka). 1 Thus, ess = (5.36) Ka where Ka = lim s 2 G (s ) H (s ) s→0

So, parabolic response of second order underdamped system is shown in Fig. 5.28.

Fig. 5.28

5.10

EFFECT OF G(s) H(s) OR TYPE OF SYSTEM ON STEADY STATE ERROR

Type of the system depends upon the sn term present in the denominator of open loop transfer function of unity feedback system. The value n is directly equal to the type of the system. It is also equal to the number of poles present at origin of s-plane in the system. Now, various types of systems are subjected to different input signals to see their effect on steady state error.

5.10.1 Type-0 System The open loop transfer function of a unity feedback system (Type-0) can be represented in time-constant form as given in equation (5.37). G (s) H (s) =

K (1 + Tz1 s) (1 + Tz 2 s) … s0 (1 + Tp1 s) (1 + Tp 2 s)…

Now, steady state error ess for various standard inputs are: For unit step input: From equation (5.34), ess =

1 1 1 = = 1 + K p 1 + lim G (s) H (s) 1 + K s→0

(5.37)

Time Response Analysis 107

This is shown in Fig. 5.29.

Fig. 5.29

Fig. 5.30

For unit ramp input: From equation (5.35), 1 1 = =∞ ess = K v lim sG (s) H (s) s→0

This is shown in Fig. 5.30. For unit parabolic input: From equation (5.36), 1 1 = =∞ ess = 2 K a lim s G (s) H (s) This is shown in Fig. 5.31.

s→0

Fig. 5.31

5.10.2 Type-1 System The open loop transfer function of a unity feedback system (Type-1) can be represented in time-constant form as given in equation (5.38). G (s) H (s) =

K (1 + Tz1 s) (1 + Tz 2 s) … s1 (1 + Tp 1 s) (1 + Tp 2 s) …

Now, steady state error ess for various standard inputs is: For unit step input: From equation (5.34), 1 1 1 = = =0 ess = 1 + K p 1 + lim G (s) H (s) 1 + ∞ s→0

(5.38)

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Automatic Control Engineering

This is shown in Fig. 5.32.

Fig. 5.32

Fig. 5.33

For unit ramp input: From equation (5.35), 1 1 1 = = ess = K v lim sG (s) H (s) K s→0

This is shown in Fig. 5.33. For unit parabolic input: From equation (5.36), 1 1 = =∞ ess = 2 K a lim s G (s) H (s) This is shown in Fig. 5.34.

s→0

Fig. 5.34

5.10.3 Type-2 System The open loop transfer function of a unity feedback system (Type-2) can be represented in time-constant form as given in equation (5.39). G (s) H (s) =

K (1 + Tz1 s) (1 + Tz 2 s) … s 2 (1 + Tp 1 s) (1 + Tp 2 s) …

Now, steady state error ess for various standard inputs is: For unit step input: From equation (5.34), 1 1 1 = = =0 ess = 1 + K p 1 + lim G (s) H (s) 1 + ∞ s→0

(5.39)

Time Response Analysis 109

This is shown in Fig. 5.35.

Fig. 5.35

Fig. 5.36

For unit ramp input: From equation (5.35), ess =

1 1 1 = = =0 K v lim s G (s) H (s) ∞ s→0

This is shown in Fig. 5.36. For unit parabolic input: From equation (5.36), ess =

1 1 1 = = K a lim s 2 G (s) H (s) K s→0

Fig. 5.37

This is shown in Fig. 5.37.

5.11

SUMMARY OF STEADY STATE ERRORS AND ERROR CONSTANTS

Steady state error for various inputs and different types of systems are summarised in Table 5.2. Table 5.2 Type of Input

Steady State Error Type-0

Type-1

Type-2

Unit step

1 1+K

0

0

Unit ramp

•

1/K

0

Unit parabolic

•

•

1/K

The three error constants i.e., Kp, Kv and Ka tell the ability of the system to eliminate the steady state error. As the type of the system becomes higher, the elimination of steady state error increases. The main limitation associated with error constants is that they give information about steady state error when inputs are the three basic types only i.e., step, ramp and parabolic.

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Automatic Control Engineering

Example 5.9

A unity feedback system is characterised by the open loop transfer 10 . Determine the static error constants. Also, determine the function G (s) = s (0.1s +1) steady state errors for unit step, unit ramp and unit acceleration inputs. Also, determine the damping factor and natural frequency of oscillations of dominant roots. Solution: Step 1: Characteristic equation is 1 + G (s) H(s) = 0 0.1 s2 + s + 10 = 0

fi

Step 2: Comparing it with general characteristic equation, wn = √10 = 3.162 rad/s and 2dwn = 1 fi δ = 0.158 Step 3: For unit step input, ess =

(Answer)

1 1 + Kp

Now,

Kp = lim G (s) H (s) = lim

So,

ess = 1/• = 0

s→0

Step 4: For unit ramp input, ess =

s→0

10 =∞ s (0.1s +1)

1 Kv

Now,

sG (s) H (s) = lim Kv = lim s→0 s→0

So,

ess = 1/10 = 0.1

Step 5: For unit parabolic input, ess =

10 = 10 (0.1s +1)

1 Ka

Now,

Ka = lim s 2 G (s) H (s) = lim

So,

ess = 1/0 = •

s→0

s→0

s 10 =0 (0.1s +1)

Example 5.10

The open loop transfer function of a unity feedback system is given 100 . Find the static error constants and steady state errors by G (s) = 2 s (s + 8) (s 2 + 6 s + 12) of the system when subjected to input given by r (t) = 5 + 2 t + 4 t2. Solution: Step 1: Position error constant, Kp = lim G (s) H (s) = lim s→0

s→0

100 =∞ s (s + 8) (s 2 + 6 s +12) 2

Time Response Analysis 111

Step 2: Velocity error constant, Kv = lim sG (s) H (s) = lim s→0

s→0

100 =∞ s (s + 8) (s 2 + 6s +12)

Step 3: Acceleration error constant, s 2 G (s) H (s) = lim Ka = lim s→0 s→0

Step 4:

100 100 = = 1.04 2 (s + 8) (s + 6 s +12) (8) (12)

r (t) = 5 + 2 t + 4 t2

Total steady state error ess of the system when subjected to r (t) is ess = Example 5.11

2 5 2 2 5 2 = = 1.923 + + + + 1 + K p K v K a 1 + ∞ ∞ 1.04

For a unity feedback system, G (s) =

(Answer)

K (1 + 2 s) , find K so that s (1 + s) (1 + 4 s) 2

steady state error of the system is 10% when subjected to input t. Solution: Step 1: When input is t i.e., unit ramp, ess = 1/Kv K (1 + 2 s ) K = =K where Kv = lim sG (s) H (s) = lim 2 s→0 s → 0 (1 + s ) (1 + 4 s ) 1 So, ess = 1/K Step 2: Also, ess = 10% = 0.1 fi K = 10

(Answer)

Example 5.12 Determine the error coefficients for a feedback system whose 100 and H (s) = (s + 5). G (s ) = s (s +10) (s + 20) Solution: 100 (s + 5) G (s) H (s) = lim =∞ Step 1: Position error constant, Kp = lim s→0 s → 0 s (s +10) (s + 20) Step 2: Velocity error constant, Kv = lim sG (s) H (s) = lim s→0

s→0

100 (s + 5) = 2.5 (s +10) (s + 20)

Step 3: Acceleration error constant, Ka = lim s 2 G (s) H (s) = lim s→0

s→0

100 s (s + 5) =0 (s +10) (s + 20)

Example 5.13 The system shown in Fig. 5.38 is a unity feedback control system with additional derivative feedback loop. (a) In the absence of derivative feedback, determine damping factor and natural frequency of oscillations. Also, determine the steady state error that will result from unit ramp input.

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(b) Determine the value of d which will increase the damping factor to 0.7. What is the change in steady state error with the addition of d in the system? (c) Discuss how steady state error in part (b) with derivative feedback can be changed to the value in part (a), if damping factor is maintained at 0.7?

Fig. 5.38

Solution: Step 1: (a) when d = 0, the characteristic equation becomes, 8 1+ =0 s (s + 2) s2 + 2 s + 8 = 0 Comparing it with equation (5.14), wn = √8 = 2.83 rad/s and 2dwn = 2 fi d = 0.353 1 1 = = For unit ramp input, ess = K v lim sG (s) H (s) fi

s→0

1 8 lim s s → 0 s (s + 2)

= 0.25

Step 2: When d is added to the system, the overall transfer function becomes, C (s ) 8 = 2 R (s) s + (2 + 8 d ) s + 8 Comparing it with equation (5.14), wn = √8 = 2.83 rad/s and 2dwn = 2 + 8d fi d = 0.245 (Answer) Now, ess =

1 1 = = K v lim s G (s) H (s) s→0

1 8 lim s 2 s → 0 s + (2 + 8 d ) s

=

1 8 s → 0 s + (2 + 8 d )

= 0.495

lim

Thus, ess is increased from 0.25 to 0.495. Step 3: To maintain ess to 0.25 with addition of d, let the system gain be adjusted from d to new value KA. So, characteristic equation becomes, s2 + (2 + KAd) s + KA = 0 (5.40) Here, 2dwn = 2 + KAd fi 2 × 0.7 × √KA = 2 + KAd Step 4: Again, ess = 0.25 =

1 = Kv

1 lim s s→0

KA s + (2 + K A d ) s 2

=

1 KA (2 + K A d )

Time Response Analysis 113

fi

0.25 KA = 2 + KAd On solving equations (5.40) and (5.41), KA = 31.36 and d = 0.186

(5.41) (Answer)

Exercise C (s ) 25 . Determine the values = R (s) s 2 + 10 s + 25 of d and wn. Obtain the unit step response of closed loop system. 2. Determine the damping factor, natural frequency of oscillations and damped frequency of oscillations of the speed control system as shown in Fig. 5.39. 1. Consider a closed loop system given by

Fig. 5.39

What is the expression for c (t) when the system is subjected to step input r (t) = 4? 3. The open loop transfer function of a unity feedback system is given by K , where K and T are constants. By what factor should the amplifier G (s ) = s (Ts + 1) gain be reduced so that the peak overshoot of the unit-step response of the system is reduced from 75% to 25%?

25 . s (s + 1) Calculate natural frequency of oscillations and damped frequency of oscillations. If it is desired to increase the damping factor to 0.7 by introducing a feedback transfer function of (2 + K) in place of unity feedback, calculate the value of K to give desired damping factor. Also calculate percentage overshoot. 5. Consider a unity feedback control system with closed loop transfer function C (s ) Ks + b . Determine the open loop transfer function. Show that the = 2 R (s) s + as + b a–K . steady state error for unit ramp input is given by b 6. A servomechanism has a maximum overshoot of 25% for unit step input and takes 2 s to reach the peak of response. Determine the damping factor and natural frequency of oscillations that will satisfy these conditions.

4. A unity feedback system having forward path transfer function G (s) =

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7. Determine the error coefficients of a unity feedback system whose open loop 60 transfer function is given by G (s) = . Also, find steady state s ( s +1) ( s + 2) (s + 3) error for r (t) = 0.1 t.

20 . s (s + 4) (s +10) 5 2 Determine error coefficients and steady state error when r (t) = 10 + 5 t + t . 2 180 and r (t) = 4. Determine 9. A unity feedback system has G (s) = s (s + 6) (i) steady state error (ii) value of new gain to reduce error by 6%. 35 (s + 4) . Find type of the system, 10. A unity feedback system having G (s) = s (s + 2) (s + 5) error constants and steady state error for ramp input having magnitude 5. 8. A unity feedback system having open loop transfer function as G (s) =

11. For a unity feedback system having G ( s ) =

K , find the following: s ( s T + 2)

(i) The factor by which the gain K should be multiplied to increase the damping factor from 0.15 to 0.6. (ii) The factor by which the time constant T should be multiplied to reduce the damping factor from 0.8 to 0.4.

Objective Type Questions 8 ( s + 3) 1. If for a control system, the Laplace transform of error e (t) is given by , s ( s + 10) then the steady state value of the error comes out to be (a) 3.6 (b) 1.8 (c) 3.2 (d) 2.4 K 2. For a unity feedback system, G (s) = s ( s + 4) , the value of K for d 0.5 is

(a) 1 (b) 16 (c) 32 (d) 64 3. For a type 1 system, ess due to step input is equal to (a) • (b) 0 (c) 0.25 (d) 0.5 4. The d of system having characteristic equation s2 + 2 s + 8 = 0 is (a) 0.353 (b) 0.330 (c) 0.300 (d) 0.250

Time Response Analysis 115

5. Laplace transform of the output response of linear system is the system transform function when the input is (a) step (b) ramp (c) impulse (d) sinusoidal 6. The ess due to ramp input for type-2 system is (a) 0 (b) • (c) non-zero number (d) constant 121 . Which of the 7. Given transfer function of a system is G (s) = 2 s + 13.2 s + 121 following characteristic does it have? (a) overdamped and settling time 1.1 s (b) underdamped and settling time 0.6 s (c) critically damped and settling time Ts 0.8 s (d) underdamped and settling time Ts 0.707 s 8. Backlash in a stable control system may cause (a) underdamping (b) overdamping (c) high level oscillations (d) low level oscillations 9. Consider the following statements with reference to the system having Kv = 1000. (i) System is stable (ii) System is type 1 (iii) Input signal is unit step. Which of the following is correct? (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 10. The d and wn of a second order system are 0.6 and 2 rad/s respectively. Which one of the following combinations gives the correct values of Tp and Ts respectively for unit step signal? (a) 3.33 s and 1.95 s (b) 1.95 s and 3.33 s (c) 1.9 s and 1.5 s (d) 1.5 s and 1.9 s 11. The open loop transfer function of a unity feedback control system is given by K G (s) = . If gain K is increased up to •, then d will tend to s ( s + 1) (a) 0 (b) 0.707 (c) 1 (d) • 12. A second order system has d and wn as dampness factor and natural frequency. The Ts for 2% tolerance band is (a) 2/d wn (b) 3/dwn (c) 4/dwn (d) dwn

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13. A system has single pole at origin. Its impulse response will be (a) constant (b) decaying exponentially (c) ramp (d) oscillating 14. What is the unit step response of a unity feedback control system having forward 80 path transfer function as G (s) = ? s ( s + 18) (a) overdamped (b) underdamped (c) critically damped (d) undamped oscillations 15. A second order system exhibit 100% overshoots. Its d is (a) 0 (b) 1 (c) less than 1 (d) greater than 1

d2y dy 16. A second order system, 2 2 + 4 + 8 y = 8 x . The d is dt dt (a) 0.1 (b) 0.25 (c) 0.333 (d) 0.5 42.25 . The s ( s + 6.5) unit step response of this system starting from test will have its maximum value at a time equal to (a) 0 (b) 0.558 s (c) 5.6 s (d) • 18. Peak overshoot of a step input response of an underdamped second order system is explicitly indicative of (a) settling time (b) rise time (c) natural frequency of oscillations (d) damping factor 17. A unity feedback system has forward transfer function equal to

Answers 1. 6. 11. 16.

(d) (a) (a) (d)

2. 7. 12. 17.

(b) (b) (c) (b)

3. 8. 13. 18.

(b) (d) (a) (d)

4. (a) 9. (a) 14. (a)

5. (c) 10. (b) 15. (a)

MATLAB Programs P1. Find the step response and impulse response of the transfer function

Program: n=[0 20];

% Numerator of transfer function

20 . s 2 + 4 s + 25

Time Response Analysis 117

d=[1 4 25]; T=tf(n,d); subplot(2,1,1);

% Denominator of transfer function % Transfer function % Divides the plot area into two subplots vertically and plots the next graph in first plot area step(T) % Plots the step response c(t) versus time in the first plot area xlabel(‘time’) % Labels the x-axis as time ylabel(‘c(t)’) % Labels the y-axis as c(t) title(‘Step response’) % Gives title to the graph subplot(2,1,2) % Divides the plot area into two subplots vertically and plots the next graph in second plot area impulse(T) % Plots the impulse response c(t) versus time in the second plot area xlabel(‘time’) % Labels the x-axis as time ylabel(‘c(t)’) % Labels the y-axis as c(t) title(‘Impulse response’) % Gives title to the graph Execution:

P2. Find the unit ramp response and unit parabolic response of the transfer function 3 s + 20 T= 2 . s + 5 s + 36 Program:

n=[3 20]; d=[1 5 36];

% Numerator of transfer function % Denominator of transfer function

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T=tf(n,d); % Transfer function % To find unit ramp response t=[0:0.5:5]; % Defining time t r=t; % Defining input r unit ramp signal as function of t subplot(2,1,1); % Divides the plot area into two subplots vertically and plots the next graph in first plot area lsim(T,r,t) % Plots the time response c(t) to input signal r w.r.t. t xlabel(‘time’) % Labels the x-axis as time title(‘Ramp response’) % Gives title to the graph % To find unit parabolic response t=[0:0.5:5]; % Defining time t r=t.*t./2; % Defining input r unit parabolic signal as function of t subplot(2,1,2); % Divides the plot area into two subplots vertically and plots the next graph in second plot area lsim(T,r,t) % Plots the time response c(t) to input signal r w.r.t. t xlabel(‘time’) % Labels the x-axis as time title(‘Parabolic response’) % Gives title to the graph Execution:

Time Response Analysis 119

P3. Find the unit ramp response of the transfer function T =

20 . s 2 + 4 s + 25

Program: n=[0 20]; d=[1 4 25]; T1=tf(n,d); T2=tf([0 1],[1 0]); T=T1*T2; step(T) xlabel(‘time’) ylabel(‘c(t)’) title(‘Ramp response’)

% % % %

Numerator of transfer function Denominator of transfer function Given transfer function T2 = 1/s

% % % %

Step of T = Ramp of T1 Labels the x-axis as time Labels the y-axis as c(t) Gives title to the graph

Execution:

P4. The open loop transfer function of a control system with unity feedback is given by 20 . Determine the damping factor, natural frequency of oscillations, G= ( s + 1) ( s + 5) damped frequency of oscillations, maximum overshoot and peak time.

Program: n=[0 20]; d=conv([1 1],[1 5]); G=tf(n,d);

% Numerator of transfer function % Denominator of transfer function % Forward path transfer function

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T=feedback(G,1) [nt,dt]=tfdata(T,’v’)

% Overall transfer function % Row vectors having coefficients of numerator and denominator of transfer function wn=sqrt(dt(3)) % Natural freqency of oscillations zeta=dt(2)/(2*wn) % Damping factor wd=wn*sqrt(1-zeta*zeta) % Damped frequency of oscillations Mp=exp(-zeta*pi/sqrt(1-zeta*zeta)) % Maximum overshoot Tp=pi/wd % Peak time Execution: nt = 0 0 20 dt = 1 6 25 wn = 5 zeta = 0.6000 wd = 4 Mp = 0.0948 Tp = 0.7854 P5. Obtain the unit step response of a second order system with zeta = 0.4 and natural frequency of oscillations = 5 rad/s.

Program: wn=5; zeta=0.4; n=[0 wn*wn]; d=[1 2*zeta*wn wn*wn]; T=tf(n,d) step(T)

% % % % % %

Natural frequency of oscillations Damping factor Numerator of transfer function Denominator of transfer function Overall transfer function Step response of given system

Time Response Analysis 121

Execution:

Transfer function: 25 -------------s^2 + 4 s + 25 P6. The open loop transfer function of a control system with unity feedback is given 10 by G = . Find the error constants and steady state error for input 3, 2 t and ( s + 2) t2/2.

Program: n=[1 10]; d=[1 2]; G=tf(n,d); Kp=dcgain(G) ess=3/(1+Kp) nv=[10 0]; dv=d; Gv=tf(nv,dv); Kv=dcgain(Gv) ess=2/Kv

% % % % % % % % % %

Numerator of G(s)H(s) Denominator of G(s)H(s) G(s)H(s) Position error constant Steady state error for input = 3 Numerator of sG(s)H(s) Denominator of sG(s)H(s) sG(s)H(s) Velocity error constant Steady state error for input = 2t

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na=[10 0]; dv=d; Gv=tf(nv,dv); Kv=dcgain(Gv) ess=2/Kv Execution: Kp = 5 ess = 0.5000 Kv = 0 ess = Inf Kv = 0 ess = Inf

% % % % %

Numerator of s2G(s)H(s) Denominator of s2G(s)H(s) s2G(s)H(s) Acceleration error constant Steady state error for input = t2/2

6

Concepts of Stability and Its Algebraic Solutions

CHAPTER

6.1

INTRODUCTION

In chapter 5, the time response of the system was studied that has two parts i.e., transient response and steady state response. Next comes the stability of the system. The study of whether the system response reaches steady state after passing through the transients is called stability study. In general, if after small changes in the input or in system parameters, the output response of the system doesn’t change much, then the system is called stable. Stability is a very important aspect for studying the transient performance and design of a control system. The main expectation of a control system is its stability in addition to meeting its main objective.

6.2

CONCEPT OF STABILITY

If the system is excited by some input signal and its response doesn’t reach steady state and goes on increasing in magnitude so that ultimately the system breaks down and becomes non-linear, then the system is said to be unstable. The mathematical model that was derived for the original linear system no longer applies in this unstable condition. So, a linear time invariant system is said to be stable if the following conditions are satisfied: (i) When the control system is excited by a bounded or finite input signal, its output signal is also bounded. This criterion is called BIBO stability (bounded input bounded output). (ii) When the input signal to the control system becomes zero, its output signal also tends towards zero, whatsoever the initial conditions may be. This criterion is called asymptotic stability. The two above said conditions are essentially equivalent to each other for a linear time invariant system. For physical realisation of these conditions; consider a linear time invariant single input single output control system having transfer function as

where n > m

xm s m + xm – 1 s m – 1 + … + x0 s 0 C (s ) = T (s ) = R (s ) yn sn + yn – 1 sn – 1 + … + y0 s0

(6.1)

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Considering zero initial conditions, c (t) = L–1 [T (s) R (s)] Now applying convolution theorem, ∞

c (t) = ∫ T ( A) r (t – A) dA

(6.2)

0

where T (A) = L–1 [ T (s)] is the impulse response of the system and A is a dummy variable. Taking absolute on both sides of equation (6.2), ∞

|c (t)| = ∫ T ( A) r (t – A) dA

(6.3)

0

As the absolute value of the integral is not greater than the integral value of the absolute of the integrand, equation (6.3) can be written as: ∞

|c (t)| £ ∫ T ( A) r (t – A) dA 0

∞

fi

|c (t)| £ ∫ T ( A) r (t – A) dA

(6.4)

0

Now, the input is bounded if and only if |T (A)| £ P1 £ • and output is bounded if and only if |c (t)| £ P2 £ •, where P1 and P2 are finite constants. So, using equation (6.4), the condition of stability becomes, c (t ) ≤ P1

∞

∫ T ( A) dA ≤ P . The system becomes 2

0

stable if and only if impulse response of T (A) is absolutely integrable i.e.,

∞

∫ T ( A) dA is 0

finite which means the area under the absolute curve of impulse response T (A) must be finite from t = 0 to t = •. The nature of T (A) depends on the poles of the transfer function of the system that are same as the roots of the characteristic equation. All possible types of poles and the corresponding nature of impulse response and hence stability is shown in Fig. 6.1. Following conclusions can be made: (i) If the poles have negative real parts, then the impulse response is bounded and gradually tends to zero, following its input. So, the system is BIBO stable and asymptotically stable. (ii) If the poles have positive real parts, then the impulse response is unbounded and goes on increasing. So, system is unstable according to both the conditions. (iii) If the poles are non-repeated and lie on the imaginary axis, then impulse response is bounded but it doesn’t tends to zero. So, system is BIBO stable but asymptotically unstable. Hence it is marginally stable. (iv) If the poles are repeated and lie on imaginary axis, then impulse response is unbounded and goes on increasing. Hence, system is unstable. Based on the above conclusions, all the sub-figures of Fig. 6.1 can be summarised into Fig. 6.2.

Concepts of Stability and Its Algebraic Solutions

125

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Fig. 6.1

Concepts of Stability and Its Algebraic Solutions

127

Im

Stable

Unstable

Re

Non-repeated— Non-repeated —Marginally stable

Repeated— Repeated —Unstable

Fig. 6.2

In more general words, if the oscillations set up in the response following an input, die out with respect to time, then the system is stable. If these oscillations go on increasing, then the system is unstable. If these oscillations are sustained, means these are neither increasing nor decreasing, the system is said to be marginally stable.

6.3

TYPES OF STABILITY

The stability of a system is classified into two categories: absolute stability and relative stability.

6.3.1 Absolute Stability This stability tells whether the system is stable or not, for all values of parameters. It is the qualitative analysis of stability. It gives the indication whether the oscillation setup in the system following an input will die out with respect to time or not. It can be determined by the location of poles of the transfer function in s-plane.

6.3.2 Relative Stability This stability tells how much stable the system is, for certain range of parameters. It is the quantitative analysis of stability. It tells how fast the oscillations will die out in the system with respect to time. It can be determined with the help of settling time of the system. It is also used to compare the systems in terms of stability. The system that has lesser settling time is more stable than the system having higher settling time.

6.4 Case 1:

NECESSARY CONDITION FOR STABILITY

Consider a characteristic equation of first order system: a0 s + a1 = 0

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Here, pole is at s = – a1/a0, same sign of a1 and a0 or positiveness of a1 and a0 ensures the pole with negative real part and hence, the system is stable. Case 2: Consider a characteristic equation of second order system: a0 s2 + a1 s + a2 = 0 Here poles are at s =

− a1 ± a12 − 4 a0 a2

. 2 a0 Again, positiveness of a2, a1 and a0 ensures the negative real part of poles and hence, the system is stable. So, it can be concluded that for the first and second order systems, positiveness of all the coefficients in characteristic equation from higher to lower degree is the necessary and sufficient condition for their stability. Case 3: Consider a characteristic equation of 3rd order system: s 3 + s2 + 2 s + 8 = 0 Here, poles are at s = – 2, 0.5 ± j √5/2 It can be seen that although all the coefficients from highest to lowest degree in the characteristic equation are present and are positive, still two poles have positive real part that make the system unstable. So, from third and higher order systems, if some of the coefficients in the characteristic equation are missing or negative, then the system is unstable. But if all the coefficients are present and are positive, system may or may not be stable. Hence the positiveness of coefficients of the characteristic equation is the necessary but not the sufficient condition for the third and higher order system to be stable. Thus, some other methods have to be discussed that deal with higher order systems.

6.5

HURWITZ STABILITY CRITERION

Consider a characteristic equation of nth order control system,

a0 sn + a1 sn – 1 + a2 sn – 2 + … + an = 0 According to Hurwitz stability criterion, “The necessary and sufficient condition for the system to be stable is that all the n determinants formed as the principal minors of the Hurwitz determinant must be positive.” Hurwitz determinant is given by equation (6.5) a1 a0 0 0

a3 a2 a1 a0

a5 a4 a3 a2

– – – –

– – – –

– – – –

– a2 n – 1 – a2 n – 2 – a2 n – 3 –

H= 0

0

a1

– – – –

0

0

0

– – – –

(6.5)

an

n×n

Concepts of Stability and Its Algebraic Solutions

129

Its principal minors are: a1 M1 = |a1|; M2 = a 0

a1 a3 a a2 ; M3 = 0 0

a3 a2 a1

a5 a4 ; º; M = |H| n a3

For the system to be stable, all Mi’s where i = 1, 2, 3, º, n must be positive. If Mn – 1 = 0, then the system is marginally stable. For higher order systems, it becomes very cumbersome to form Hurwitz determinant. A new method is developed by E.J. Routh popularly known as Routh’s criterion. Example 6.1 Consider a system having characteristic equation as s4 + 5 s3 + 9 s2 + 8 s + 2 = 0. Determine its stability using Hurwitz criterion. Solution: Step 1: Characteristic equation is s4 + 5 s3 + 9 s2 + 8 s + 2 = 0 5 1 Hurwitz determinant is H = 0 0

Step 2:

8 6 5 1

0 2 8 6

0 0 0 2 4×4

M1 = |5| = 5 M2 =

5 8 = 22 1 6

5 8 0 M3 = 1 6 2 = 5(38) – 8(8) = 190 – 64 = 126 0 5 8 5 1 M4 = 0 0

8 6 5 1

0 2 8 6

0 0 = 252 0 2

Step 3: As all Mi’ s; i = 1, 2, 3, 4 are positive, system is stable.

6.6

(Answer)

ROUTH’S STABILITY CRITERION

This method is also known as Routh’s Hurwitz criterion or R-H criterion. This method first describes the arrangement of coefficients of characteristic equation into a particular array known as Routh’s array, which is described as follows: Consider an nth order characteristic equation: a0 sn + a1 sn – 1 + a2sn – 2 + º + an – 1 s1 + ans 0 = 0

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So, Routh’s array becomes sn n–1

s sn – 2 sn – 3 . . s s0

a0 a1 b1 c1 . . . .

a2 a3 b2 c2 . . . .

a4 … a5 … b3 …

The first two rows are made using coefficients of characteristic equation. Rest of the elements of Routh’s array are computed as follows: a1 a2 − a0 a3 a1 a4 − a0 a5 ; b2 = ;º b1 = a1 a1 b1 a3 − a1 b2 b1 a5 − a1 b3 ; c2 = ;º b1 b1 It should be noted that all the coefficients from highest degree to the lowest in the characteristic equation must be taken into account. Any missing coefficient can be regarded as zero. Also, any row of Routh’s array can be divided by a common positive constant so as to simplify the calculations further. The Routh’s array is continued up to the row of s0. According to Routh’s criterion, “For a system to be stable, the necessary and sufficient condition is that each term in the first column of Routh’s array must have positiveness i.e., all the terms must be of same sign. If this condition is not met, the system is said to be unstable. The number of poles due to which the system becomes unstable or the number of poles that lie on R.H.S. of the s-plane is equal to the number of sign changes in the first column of Routh’s array”. It is very important to note that Hurwitz criterion and Routh’s Hurwitz criterion are equivalent to each other. a1 a3 a1 a2 − a0 a3 a0 a2 M = = 2 in Hurwitz criterion. In R-H criterion, b1= a1 a1 M1

Similarly,

c1 =

So, b1 is positive iff both M2 and M1 are positive. Again, c1 =

b1 a3 − a1 b2 M 3 = in Hurwitz criterion. b1 M2

So, condition of positiveness of Hurwitz determinant is equivalent to the condition of positiveness of elements in the first column of Routh’s array. Example 6.2 Consider the characteristic equation of Example 6.1. Determine its stability using Routh’s Hurwitz criterion.

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131

Solution: Step 1: Characteristic equation is: s4 + 5s3 + 6s2 + 8s + 2 = 0 Routh’s array becomes s4 s

3

s2 s1 s0

1 5 5×6–1×8 = 4.4 5 4.4 × 8 – 5 × 2 = 5.73 4.4 5.73 × 2 – 4.4 × 0 =2 5.73

6 2 8 0 5×2–1×0 =2 0 5 4.4 × 0 – 5 × 0 =0 4.4 0

Step 2: As all the terms in the first column of Routh’s array are positive, the system is stable. Example 6.3 For the system having characteristic equation as 5 s4 + 16 s3 + 5 s2 + 8 s + 2 = 0, discuss stability using R-H criterion. Solution: Step 1: Characteristic equation is 5 s4 + 16 s3 + 5 s2 + 8 s + 2 = 0 s4 5 s 3 16 s 2 2.5 s1 – 3 s0 2

5 2 s4 5 8 0 s3 2 2 = s 2 2.5 0 s1 – 0.6 s0 2

5 2 1 0 2 0 0

In order to simplify the calculation, the second row is divided by a positive constant 8. Step 2: The fourth term in first column of Routh’s array is negative. Hence the system is unstable. Step 3: There are two sign changes in the first column. First sign change is from + 2.5 to – 0.6 and the second sign change is from – 0.6 to + 2. So, there must be two poles in the R.H.S. of the s-plane that are making this system unstable. (Answer)

6.7

SPECIAL CASES OF ROUTH’S HURWITZ CRITERION

Sometimes, while applying Routh’s Hurwitz criterion to determine the stability of a control system, some problems are encountered, that results in breaking this stability test. These difficulties can be of the following kinds: Problem 1: Case 1: When the first element in any row of the Routh’s array is zero while all other elements in that row are non-zero.

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Due to this zero term, the elements in the next row become infinite and Routh’s Hurwitz criterion test breaks down. This problem can be overcome using two methods: Method 1: Replace the zero term by a small positive number t and proceed to evaluate rest of the Routh’s array in terms of t. Then examine the sign change in the first column by putting t → 0. Method 2: Replace variable s by another variable 1/z and modify the original characteristic equation. Now apply Routh’s Hurwitz criteria on the modified characteristic equation and examine the stability. The number of z-roots in RHS of z-plane is same as the number of s-roots in RHS of s-plane. Example 6.4 By using Routh’s Hurwitz criteria, determine stability of the system having characteristic equation as s 5 + 2 s4 + 6 s3 + 12 s2 + 6 s + 10 = 0. Solution: Step 1: Characteristic equation is s5 + 2 s4 + 6 s3 + 12 s2 + 6 s + 10 = 0 Routh’s array is s5 1 6 6 s 4 2 12 10 s3 0 1 0 s2 s1 s0

Step 2: As the first term in the third row is zero, the next term becomes infinity. So, replace this zero by a small positive number t. The new Routh’s array becomes s5 s4 s3 s2 s1 s0

1 1 τ 6τ – 1 τ – 5 τ2 + 6 τ – 1 6τ – 1 5

Step 3: To examine stability, put t → 0

6 6 1

6 5 0

5

0

0

0

0

0

6t − 1 =−∞ t − 5t 2 + 6 t − 1 And first term in fifth row is lim =+1 t→0 6t − 1

First term in fourth row becomes lim t→0

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Step 4: New Routh’s array is 6 6 s5 1 4 2 12 10 s 3 τ 1 0 s 2 s –∞ 5 0 0 0 s1 1 0 5 0 0 s

Step 5: System is unstable with two sign changes. Hence two poles lie in RHS of s-plane. Example 6.5 Determine the stability of the system having characteristic equation given in Example 6.4. Use method 2 of case 1. Solution: Step 1: Characteristic equation is s5 + 2 s4 + 6 s3 + 12 s2 + 6 s + 10 = 0 5

Replace s by fi

4

3

2

1 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ , we get ⎜ ⎟ + 2 ⎜ ⎟ + 6 ⎜ ⎟ + 12 ⎜ ⎟ + 6 ⎜ ⎟ + 10 = 0 ⎝ z⎠ ⎝ z⎠ ⎝ z⎠ ⎝ z⎠ ⎝ z⎠ z 5 4 3 2 10 z + 6 z + 12 z + 6 z + 2 z + 1 = 0

Step 2: Routh’s array is z5 z4 z3 z2 z1 z0

10 6 1 5 – 0.034 1

12 6 0.166 1 0 0

2 1 0 0 0 0

Step 3: As first term in fifth row is negative, system is unstable. Sign changes twice, hence two poles lie in RHS of z-plane. (Answer) Example 6.6 Apply Routh’s Hurwitz criteria to check the stability of system having characteristic equation s5 + s4 + 2 s3 + 2 s2 + 6 s + 5 = 0 Solution: Step 1: Routh’s array is s5 s4 s3 s2 s1 s0

1 2 6 1 2 5 0 1 0

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As first term in third row of Routh’s array is zero, so rest of the rows in Routh’s array cannot be determined and the test fails. Step 2: Method 1: Replace zero by a small positive number t and determine the Routh’s array in terms of t. 2 6 s5 1 4 2 5 s 1 1 0 s3 t 2t – 1 5 0 s2 t – 5 t2 + 2 t – 1 s1 0 0 2t – 1 0 0 s0 5 1 = –• Step 3: To examine stability put t → 0. Therefore, first term in fourth row = – 0 –1 = 1. and first term in fifth row = –1 The modified Routh’s array becomes s5 1 s4 1 s3 t s2 – • s1 1 s0 5

2 2 1 5 0 0

6 5 0 0 0 0

Step 4: As the fourth term in first column of Routh’s array is negative, system is unstable. Step 5: As sign changes twice in first column of Routh’s array, out of total five poles, there are two poles which lie in RHS of s-plane, due to which the system is unstable. Step 6: Method 2: Replace variable ‘s’ by another variable ‘1/z’ in the original characteristic equation, so new characteristic equation becomes 5

4

3

2

Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ ÁË ˜¯ + ÁË ˜¯ + 2 ÁË ˜¯ + 2 ÁË ˜¯ + 6 ÁË ˜¯ + 5 = 0 z z z z z

fi 5 z5 + 6 z4 + 2 z3 + 2 z2 + z + 1 = 0 Step 7: Routh’s array is z5 z4 z3 z2 z1 z0

5 2 6 2 1 /3 1 /6 –1 1 1/2 0 1 0

1 1 0 0 0 0

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Step 8: As the fourth term in first column of Routh’s array is negative, system is unstable. Step 9: As sign changes twice in first column of Routh’s array, out of total five poles, there are two poles which lie in RHS of s-plane, due to which system is unstable. (Answer) Problem 2: Case 2: When all the terms in any row of Routh’s array is zero. Due to this zero row, next rows of Routh’s array are impossible to found out and thus the test fails. This zero row indicates that the system has at least one pair of poles that lie radially opposite to each other and are equidistant from origin. In this case, auxiliary polynomial A (s) needs to be found out, which can be formed using the elements of the row just above the rows of all zeroes. The order of auxiliary polynomial is always positive. The solution of this auxiliary polynomial A (s) gives the number and location of poles that are located radially opposite to each other. Rest of the Routh’s array is completed by using the coefficients of first order of derivative of A (s). Example 6.7 Apply Routh’s Hurwitz criteria to investigate the stability of system having characteristic equation 2 s5 + 2 s4 + 8 s3 + 8 s2 + 4 s + 4 = 0 Solution: Step 1: Routh’s array is s5 2 8 4 s4 2 8 4 s3 0 0 0 s2 s1 s0

As all the terms in third row are zero, the test fails. Step 2: A (s) is formed using the terms in second row, A (s) = s4 + 4 s2 + 2 s0 Step3: Take first order derivative of A (s) A¢(s) = 4 s3 + 8 s Step 4: Now replace the zero elements row by coefficient of A¢(s) in Routh’s array, and proceed further. s5 s4 s3 s2 s1 s0

1 1 4 2 1 2

4 2 4 2 8 0 2 0 0

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Step 5: As all the terms in first column are positive, system may or may not be stable. To check stability further, solve A (s). – 4 ± 16 – 8 – 4 ± 2 2 = – 3.415, – 0.585 s 4 + 4 s 2 + 2 = 0 fi x2 + 4 x + 2 = 0 fi x = = 2 2 Thus, s = ± j 1.848; ± j 0.765 As the roots of A (s) are non-repeated and imaginary, the system is marginally stable. Important Note: In special Case II, when all the terms in any row are zero, then A (s) needs to be found out. Routh’s array is modified using coefficients of A¢(s). If in the modified Routh’s array, any term in first column is negative, the system is unstable and test is completed. But if all the terms in first column are positive, system may or may not be stable. To check further, A(s) is solved and stability is judged on the basis of roots of A(s). Example 6.8 Investigate the stability using Routh’s Hurwitz criteria for the following characteristic equation s6 + 3 s5 + 5 s4 + 9 s3 + 8 s2 + 6 s + 4 = 0 Solution: Step 1: Routh’s Array is s6 s5 s4 s3 s2 s1 s0

1 3 2 0

5 8 4 9 6 6 4 0

Step 2: As all the terms in 4th row are zero, A (s) needs to be found out as A (s) = s4 + 3 s2 +2 s0 Step 3: A′ (s) = 4 s3 + 6 s So, modified Routh’s array becomes s6 s5 s4 s3 s2 s1 s0

1 3 1 4 1.5 0.333 2

5 9 3 6 2 0 0

8 4 6 2 0 0

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Step 4: As all the terms in first column of the modified Routh’s Array are positive, system may not be stable. Step 5: Now, A (s) = s4 + 3 s2 + 2 = 0 On solving, s = ± j √2; ± j 1 As the roots are non-repeated imaginary, the system is marginally stable. (Answer)

6.8

APPLICATIONS OF ROUTH’S HURWITZ CRITERION

It is seen that Routh’s Hurwitz criterion is used to determine the stability of a control system, having constant parameters. If some of the parameters are variable in a control system, then Routh’s Hurwitz criterion can be used to find the condition of stability in terms of those variable parameters. Consider a feedback control system as shown in Fig. 6.3.

Fig. 6.3

Here characteristic equation is 1 + KG (s) H (s) = 0. To determine the range of K for stability, Routh’s array is constructed in terms of K. Then the range of K for stability can be obtained so that no sign changes occur in the first column of Routh’s array. Example 6.9

The open loop transfer function of feedback system is given by G (s) K . Determine the maximum value of K for the system to be H (s) = 4 s + 6 s 3 + 11 s 2 + 6 s stable using Routh’s Hurwitz criterion. Solution: Step 1: Characteristic equation is 1 + G (s) H (s) = 0 fi s 4 + 6 s3 + 11 s2 + 6 s + K = 0 Step 2: Routh’s array becomes s4 1 s3 6 s 2 10 s1 60 – 6 K 10 s0 K

11 K 6 K 0

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Step 3: For the system to be stable, each term in the first column of Routh’s array must be positive. 60 – 6 K > 0 fi K < 10 and K > 0 10 Step 4: Thus, maximum value of K for which the system is stable is Kmax = 10 (Answer)

So,

Example 6.10 The characteristic equation of the feedback control system is s4 + 20 s3 + 15 s2 + 2 s + K = 0. Comment on the stability of system in terms of K. Solution: Step 1: Routh’s array is 1 20 14.9 14.9 – 10 14.9

15 2

Step 2: System is stable when all the terms in first column of Routh’s array are positive. 14.9 – 10 K > 0 fi K < 1.49 and K > 0 14.9 Thus, range of K for the system to be stable is 1.49 > K > 0 Step 3: System is unstable when K > 1.49 and K < 0 Step 4: System is marginally stable at K = 1.49

So,

(Answer) (Answer) (Answer)

Example 6.11 A feedback control system is shown in Fig. 6.4. Determine K so that system becomes stable.

Fig. 6.4

Solution: Step 1: Characteristic equation is 1 + G (s) H (s) = 0 K =0 s (s + 7) (s + 1)

fi

1+

fi

s 3 + 8 s2 + 7 s + K = 0

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139

Step 2: Routh’s array is s3 s

2

s1 s0

1 8 56 – K 8 K

7 K 0

Step 3: For the system to be stable, all the terms in first column of Routh’s array must be positive. 56 – K > 0 fi K < 56 and K > 0 8 Step 4: Thus, range of K for stable system is 56 > K > 0

So,

(Answer)

Example 6.12 Determine the maximum value of K for which the system is stable having characteristic equation as s4 + 3 s3 + 3 s2 + 2 s + K = 0. Also determine the corresponding frequency of oscillations. Solution: Step 1: Routh’s array becomes, 1 s4 3 s3 7 /3 s2 14 / 3 – 3 K s1 7 /3 s0 K

3 K 2 K 0

14 / 3 – 3 K > 0 fi K < 1.555 and K > 0 7 /3 So, range of K for stability becomes 1.555 > K > 0

Step 2: For the system to be stable,

Step 3: Now, maximum value of K for the system to be stable is Kmax = 1.555 (Answer) Step 4: For this maximum value of K, fourth row in Routh’s array becomes zero. So, auxiliary polynomial A (s) = 7/3 s2 + 1.555 = 0 fi s = ± j 0.816 Hence, frequency of oscillations = 0.816 rad/s. (Answer) Ke– 2 s , Example 6.13 A closed loop system whose transfer function is G (s) H (s) = s (s + 1) determine the maximum value of K for stability. Solution: Step 1: Characteristic equation is 1 + G (s) H (s) = 0 fi

s (s + 1) + Ke – 2 s = 0

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Step 2: Now, e –2 s = 1 –

2 s (2 s )2 + –º 1! 2!

Neglecting second and higher order terms, e –2 s ª (1 – 2 s) Therefore, characteristic equation becomes s (s + 1) + K (1 – 2 s) = 0 s2 + (1 – 2 K) s + K = 0

fi

Step 3: Routh’s array is s2 K 1 1 s 1 – 2K 0 s0 K 0

Step 4: For system to be stable, all the elements of first column should be non-zero. Therefore, (1 – 2 K) > 0 fi K < 0.5 and K > 0 Step 5: Maximum value of K for stability is K = 0.5.

(Answer)

Example 6.14 Find out Kmax and frequency of oscillations of the system given by characteristic equation s4 + 22 s3 + 10 s2 + s + K = 0. Solution: Step 1: Routh’s array is s4 1 s 3 22 s 2 9.95 9.95 – 22 K s1 22 s0 K

10 K 1 0 K 0 0

0

0

0

Step 2: For system to be stable, all elements of first column of Routh’s array should be non-zero. 9.95 – 22 K > 0 fi K < 0.453 and K > 0 22 So, maximum K = 0.453

Therefore, Step 3:

Step 4: For this value of K, fourth row of Routh’s array becomes zero. Therefore,

A (s) = 9.95 s2 + 0.453

Step 5: To find roots of A (s), put A (s) = 0 fi 9.95 s2 + 0.453 fi s = ± j 0.213 Therefore, frequency of oscillations is 0.213 rad/s.

(Answer)

Concepts of Stability and Its Algebraic Solutions

6.9

141

RELATIVE STABILITY ANALYSIS

As already discussed, the relative stability analysis is a quantitative analysis and can be estimated with the help of settling time of the system. Settling time is inversely proportional to the real part of the roots of the system, so relative stability can be specified by checking that all the roots of a characteristic equation are more negative than a certain value, say l. So, all the poles of the system must lie on the LHS of line s = – l (l > 0) in s-plane. In this condition, the original characteristic equation is modified by shifting s = z – l in the original characteristic equation. This criterion is shown in Fig 6.5.

Fig. 6.5

The system is said to be relatively stable with respect to s = – l, if all the poles lie on the left hand side of the z-plane. 2 and H (s) = 1. Determine its relative Example 6.15 A system has G (s) = s (s + 1) (s + 2) stability about the line s = –1. Solution: Step 1: Original characteristic equation is 1 + G (s) H (s) = 0 fi s (s + 1) (s + 2) + 2 = 0 fi (s2 + s) (s + 2) + 2 = 0 fi s3 + 3s2 + 2s + 2 = 0 Step 2: Replace s by (z – 1), (z – 1)3 + 3 (z – 1)2 + 2 (z – 1) + 2 = 0 fi z3 – 3 z2 + 3 z – 1 + 3 z2 + 3 – 6 z + 2 z – 2 + 2 = 0 fi z3 – z + 2 = 0 fi z3 + 0z2 – z + 2 = 0 Step 3: Routh’s array is z3 1 – 1 0 z2 0 2 0 z1 z0

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Step 4: As first term in second row is zero, replace it by small positive number t. Now Routh’s array becomes, –1 0 z3 1 2 0 z2 t –t – 2 0 0 z1 t z0 2 0 0 2 Step 5: For checking stability, put t → 0, first term in third row is – = – • 0 Therefore, Routh’s array reduces to z3 1 –1 0 z2 t 2 0 z1 – • 2 z0

0 0

0 0

Step 6: As third term in first column of Routh’s array is negative, system is unstable about line s = – 1. Example 6.16 Check whether the system is stable with respect to line s = – 2, having characteristic equation as s3 + 7 s2 + 25 s + 39 = 0. Solution: Step 1: Put s = (z – 2) in the original characteristic equation, (z – 2)3 + 7 (z – 2)2 + 25 (z – 2) + 39 = 0 fi z3 – 8 – 6 z2 + 12 z + 7 z2 + 28 – 28 z + 25 z + 39 = 0 fi z3 + z2 + 9 z + 9 = 0 Step 2: Routh’s array is z3 1 9 0 z2 1 9 0 z1 0 0 0 z0 0 0 0 Step 3: As all the terms in third row of Routh’s array are zero, A (s) has to be found out. A (s) = z2 + 9 Step 4:

A¢(s) = 2 z

Step 5: New Routh’s array becomes z3 z2 z1 z0

1 1 2 9

9 9 0 0

0 0 0 0

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143

Step 6: As all the terms in the first column are positive, system may or may not be stable with respect to s = – 2. Step 7: To check further, put A (s) = 0 fi z2 + 9 = 0 fi z = ± j 3. Roots of A (s) are non-repeating imaginary, system is marginally stable with respect to s = – 2. Example 6.17 Find the range of K such that characteristic equation s3 + 3 (K + 1) s2 + (7 K + 5) s + (4 K + 7) = 0 has roots more negative than s = – 1. Solution: Step 1: Put s = (z – 1) in the characteristic equation, we get ( z - 1)3 + 3 (K + 1) ( z - 1) 2 + (7 K + 5) ( z - 1) + (4 K + 7) = 0

fi z3 - 1 - 3 z 2 + 3 z + 3 K z 2 + 3 z 2 + 3 K + 3 - 6 K z - 6 z + 7 z + 5 z - 7 k - 5 + 4 K + 7 = 0 fi z3 + ( - 3 + 3 K + 3) z 2 + (3 - 6 K - 6 + 7 K + 5) + ( - 1 + 3 K + 3 - 7 K - 5 + 4 K + 7) = 0 fi z3 + 3 K z 2 + (2 + K ) z + 4 = 0 Step 2: Routh’s array is z3 z

2

z1 z0

2+K 0 4 0

1 3K 6K + 3K2 – 4 3K 4

0

0

0

0

Step 3: For system to be stable with respect to s = –1, all the terms in first column of Routh’s array should be non-zero. Therefore, 3 K > 0 fi K > 0 and and fi fi

6 K + 3K 2 - 4 >0 3K - 6 ± 36 + 48 6 K > – 2.53, – 0.528

K>

Step 4: Range of K for the system to have roots more negative than s = – 1 is • > K > 0.528.

Exercise 1. Using Routh’s Hurwitz criterion, find the stability of the system having following characteristic equations. If the system is unstable, find the total number of poles that lie in right hand side of s-plane.

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(a) s4 + 2 s3 + s2 + 4 s + 2 = 0 (b) s5 + s4 + 3 s3 + 9 s2 + 16 s + 10 = 0 (c) s4 + 2 s3 + 8 s2 + 4 s + 3 = 0 (d) 8 s5 + 4 s4 + 7 s3 + 5 s2 + s + 11 = 0 2. With the help of Hurwitz criterion, find stability for the following characteristic equations: (a) s4 + 2 s3 + s2 + 4 s + 2 = 0 (b) s5 + s4 + 3 s3 + 9 s2 + 16 s + 10 = 0 (c) s4 + 2 s3 + 8 s2 + 4 s + 3 = 0 (d) 8 s5 + 4 s4 + 7 s3 + 5 s2 + s + 11 = 0 3. Find the value of K for the system to be stable given by following characteristic equation s4 + 4 s3 + 14 s2 + 26 s + K = 0 4. The open loop transfer function of unity feedback control system is given by K . Apply the Routh’s Hurwitz criterion to discuss G (s) = (s + 2) (s + 4) (s 2 + 6 s + 25)

5. 6.

7.

8.

the stability of system in terms of K. Determine the value of K that will cause sustained oscillations in the system. Also determine the frequency of oscillations. s+9 . The open loop transfer function of unity feedback system is G (s) = 2 s + 8s + 9 Determine its stability. A unity feedback control system is having open loop transfer function as K (s + 13) G (s) = . Using Routh’s Hurwitz criterion, calculate the range of K s (s + 3) (s + 7) for the system to be stable. Ke -2 s . A feedback system has open loop transfer function as G (s) = s (s 2 + 5 s + 9) Determine K for the system to be stable using Routh’s Hurwitz criterion. K . The open loop transfer function of unity feedback system is G (s) = 2 s (s + 8 s + 10) Using Routh’s Hurwitz criterion, determine the range of K for the system to be more stable than the line s = – 1.

Objective Type Questions 1. If poles of a control system lie on the RHS of s-plane, the system will be (a) unstable (b) stable (c) marginally stable (d) critically stable

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145

2. If all the poles of a control system lie on the imaginary axis of s-plane, the system will be (a) unstable (b) marginally stable (c) stable (d) either (a) or (b) 3. If all the poles of a control system lie on the LHS of s-plane, the system will be (a) unstable (b) stable (c) marginally stable (d) critically stable 4. Routh’s Hurwitz criterion for determining stability of a control system is (a) frequency domain method (b) time domain method (c) both (d) none 5. Which of the following system is stable? (a) As4 + Bs3 + Cs + D = 0 (b) As4 + Bs3 + Cs2 + Ds + E = 0 (c) – As4 + Bs3 + Cs + D = 0 (d) – As4 + Bs3 + Cs2 + Ds – E = 0 6. The first column of Routh’s array is 3, 5, 12, 32. The system is (a) stable (b) unstable (c) marginally stable (d) critically stable 7. The first column of Routh’s array is 3, 5, – 12, 32. The system is (a) stable (b) unstable (c) marginally stable (d) critically stable 8. The Routh’s Hurwitz criterion cannot be applied when any coefficient of the characteristic equation is (a) exponential and sinusoidal (b) exponential (c) sinusoidal (d) complex 9. Hurwitz criterion and Routh’s Hurwitz criterion are analogous to each other. (a) True (b) False

Answers 1. (a) 6. (a)

2. (d) 7. (b)

3. (b) 8. (d)

4. (d) 9. (a)

5. (b)

7

Root Locus Technique

CHAPTER

7.1

INTRODUCTION

The importance of stability of control system has been already discussed in the previous chapter. Almost every feedback control system becomes unstable due to the nature of feedback. But an unstable system is not required as it cannot perform the control task that is required of it. So, stability analysis becomes one of the foremost analyses of control system. One method to find stability is already discussed in the previous chapter that is Routh’s Hurwitz criterion. This method determines stability in terms of location of roots of characteristic equation which are also poles of closed loop transfer function of the control system. This stability is absolute stability. This method has many shortcomings that leads to the development of a more valuable method i.e., root locus technique. This new method of analyzing stability has many advantages like: (i) It is able to tell the relative stability also (Tp , Tr , Mp , Ts , etc.) (ii) It is effective for characteristic equation having degree three or higher. (iii) It is a graphical technique so it is less time consuming and less laborious. (iv) It gives the sensitivity of roots towards the parameter that is being varied over the complete range. Root locus technique is a graphical method that plots the locus of roots of transfer function in s-plane when one of the system parameters is varied over the complete range i.e., from 0 to •. It is a powerful graphical technique that was first introduced by W.R. Evans.

7.2

CONCEPT OF ROOT LOCUS: MAP CRITERION AND ANGLE CRITERION

Consider a simplified second order control system in Fig. 7.1. In this system, K is gain constant and a is another constant of the system. The overall transfer function of the system is given by equation (7.1)

C (s ) K = 2 R (s) s + as + K

(7.1)

Root Locus Technique 147 K s ( s + a)

+

R(s)

–

C(s)

Fig. 7.1

The roots of characteristic equation are given by equation (7.2) 2

s1, s2 = –

a Ê aˆ ± Á ˜ -K Ë 2¯ 2

(7.2)

It is clear from equation (7.2) that the values of roots of characteristic equation are dependent upon both a and K. Now if a is varied for complete range of values i.e., from 0 to •, then for each value of K, one set of roots is obtained. When all such locations are plotted in s-plane and joined, root locus is obtained. This method is direct and easier for characteristic equation having lesser degree. But when degree becomes 3 or higher, this direct method becomes tedious and time consuming. So, the method developed by Evans is described as follows: The general characteristic equation can be written as: 1 + G (s) H (s) = 0 fi

G (s) H (s) = – 1 As s is a complex variable, equation (7.3) can be written in polar form as |G (s) H (s)| = 1

and

(7.3) (7.4)

– G (s) H (s) = ± (2q + 1) 180∞; q = 0,1, 2, •

(7.5)

So, G (s) H (s) = – 1 gives those values of s that has magnitude equal to 1 and angle equal to ± 180°, ± 540°, ± 900°,º, ± (2q + 1) 180°, q = 0, 1, 2, º. Equation (7.4) is referred as magnitude criterion and equation (7.5) is referred as angle criterion. Root Locus can be quickly drawn by using these two criterion. Also, any point that satisfies any of these two criterion, lies on the root locus.

7.3

CONSTRUCTION RULES OF ROOT LOCUS

Consider a general form of characteristic equation as 1 + G (s) H (s) = 0. The characteristic equation can be represented in factor form as given by equation (7.6), m

1 + G (s) H (s) = 1 +

K ’ (s + zi ) i =1

(7.6)

n

’ (s + p ) j =1

j

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where K is gain parameter of the system, m is number of zeroes of the open loop transfer function G (s) H (s) and n is the number of poles of open loop transfer function. Also n ≥ m for all physically realizable systems. Some rules have been developed for making a quick and approximate sketch of root locus using equation (7.6). Rule 1: The root locus is symmetrical about real s-axis It can be justified by knowing that the roots of open loop transfer function are either real or complex conjugate pairs or combination of both. Root locus is the line joining all these roots, so obviously it becomes symmetric about real axis. Rule 2: As one of the system parameters (here K) is varied from 0 to •, the number of branches in the root locus is equal to n. So, each branch originates from each open loop pole at K = 0. Out of these n branches, m branches will terminate at each open loop zero at K = • and remaining (n – m) branches will terminate on • at K = •. Equation (7.6) can be rewritten as: m

n

i =1

j =1

K ’ (s + zi ) + ’ (s + p j ) = 0 When K = 0, roots become s = – pj ; j = 1, 2, 3, º, n. So, the root locus branches start from open loop poles when K = 0. Again, equation (7.6) can be rewritten as n

m

j =1

i =1

(1 / K ) ’ (s + p j ) + ’ (s + zi ) = 0 When K = •, roots become s = zi ; i = 1, 2, 3, º, m. So, the root locus branches terminate at open loop zeros when K = •. The remaining (n–m) branches (because n ≥ m) terminate at •. Rule 3: Every point on real axis can be checked for its existence on root locus. Any point lies on root locus iff the number of open loop poles plus open loop zeros on the real axis to the R.H.S. of this point is odd. When any point on real axis is joined by the phasors to all open loop poles and zeros, it is seen that: (i) Angle contribution of each pole and zero on real axis to the L.H.S. of this point is 0° (ii) Angle contribution of each pole and zero on real axis to the R.H.S. of this point is 180° (iii) Net angle contribution of every complex conjugate pair to this point is always zero (+ q and – q) So, angle criterion using equation (7.5) can be written as –G (s) H (s) = ± (2q + 1) 180°; q = 0, 1, 2, º = (nr + mr) 180° where

nr = number of open loop poles on R.H.S. to that point on real axis. mr = number of open loop zeros on R.H.S. to that point on real axis.

(7.7)

Root Locus Technique 149

From equation (7.7), (2q + 1) = (nr + mr) Now, the factor (2q + 1) is always odd. Hence (nr + mr) must be odd and hence the rule is justified. It is also to be noted that this rule divides the real axis into a number of on-locus and not-on-locus segments and these segments must alternate on real axis. Example 7.1 For a unity feedback system, open loop transfer function is K . Draw the first three rules for sketching the root locus. G (s) = s (s + 1) (s + 3) Solution:

K s (s + 1) (s + 3) There are no open loop zeros. So, m = 0 Open loop poles are at s = 0, – 1, – 3. So, n = 3

Step 1:

G (s) =

Step 2: The open poles and zeros are plotted in s-plane as shown in Fig. 7.2. Im

–3

–1

Re

Fig. 7.2

Step 3: Rule 1: As all the poles lie on real axis, root locus will be symmetric about real axis. Step 4: Rule 2: There are three branches in root locus each originating from K = 0 at 0, – 1, – 3. All the three branches will terminate at K = •, on • (As n – m = 3). Step 5: Rule 3: The segments on real axis between 0 and –1 and –3 and • lie on root locus. So, these are on locus segments while the segments between – 1 and – 3 are not on locus segments. It can be verified as: for any point between – 3 and •, the number of poles and zeros on its R.H.S. is three i.e., odd, so this segment is on-locus segment. Rule 4: The (n – m) branches that terminates on •, do so along particular directions given by straight line asymptotes. These angles are given by equation (7.8).

(2 + 1) 180° ; q = 0, 1, 2, º (n – m – 1 (7.8) ( n – m) Consider a point on root locus that tends towards • (so that point is very far from poles and zeros) and join this point to all other poles and zeros through phasors. Essentially, the angle between each pole and zero to this point will be same, say F. So, for this point, fa =

– G (s) H (s) = (n – m) F

(7.9)

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Also, angle criterion for this point is – G (s) H (s) = ± (2q + 1) 180° Equating equations (7.9) and (7.10), (n – m) F = ± (2q + 1) 180° fi

fa =

± (2q + 1) 180° ( n – m)

(7.10)

(7.11)

So, the angle of asymptotes is given by equation (7.11). Now, these asymptotes are obviously (n – m). As the value of q starts from 0, so the last value of q will be (n – m – 1). Hence, rule 4 as given by equation (7.8) is justified. Rule 5: The asymptotes cross the real axis at one common point called centroid (sa) which is given by equation (7.12). sa =

 Real part of poles –  Real part of zeros n–m

Equation (7.6) can be rewritten as G (s) H (s) = G (s) H (s) = fi

(7.12)

K (s + z1 ) (s + z2 ) º (s + zm ) ;n≥m (s + p1 ) (s + p2 ) º (s + pn ) K m È n–m Ê ˘ ˆ + Á Â p j – Â zi ˜ sn – m – 1 + º˙ Ís Ëj=1 ¯ i=1 ÎÍ ˚˙ n

Ignoring higher powers of s in denominator as s → •,

K

G (s) H (s) =

m È n–m Ê ˘ ˆ + Á Â p j – Â zi ˜ sn – m + 1 ˙ Ís Ëj=1 ¯ i=1 ÎÍ ˚˙ Now consider a new function,

P (s) =

n

[s

1 + (n - m) s a sn – m –1 + º]

n–m

(7.13)

(7.14)

sa is selected such that G (s) H (s) behaves in the same manner as P (s) for s → •. So, branches of root locus of 1 + G (s) H (s) = 0 that tend towards •, approach the straight line root locus branches of 1 + P (s) = 0. From equation (7.13) and equation (7.14), n

(n – m) sa =

Â

j=1

m

p j – Â zi i=1

n

m

 p – z j

fi

sa =

j=1

i

i=1

( n – m)

(7.15)

Root Locus Technique 151

As complex poles and zeros always exist in conjugate pairs, sa is always real. Equation (7.15) can be rewritten as n

m

 (– p ) –  (– z ) j

– sa = fi

– sa =

j=1

i

i=1

( n – m)

 Real part of poles –  Real part of zeros

n–m sa can lie either on locus segment or not on locus segment of real axis.

Example 7.2 Consider the open loop transfer function of Example 7.1. Find rule 4 and rule 5 for drawing root locus. Solution: Step 1: Rule 4: Three branches will tend towards • along asymptotes whose angles are given by ± 180∞ (2q + 1) ; q = 0, 1, 2, º, (n – m – 1) fa = n-m fi fa = 180°/3; 3* 180°/3, 5* 180°/3 = 60°, 180°, 300° Step 2: Rule 5: The point of intersection of asymptotes with real axis is given by centroid sa . Â Real part of poles – Â Real part of zeros sa = n–m (0 – 1 – 3) – (0) = - 1.33 = 3 Step 3: Rules 4 and 5: These are shown in Fig. 7.3. A1, A2, A3 are three asymptotes at angles 60°, 180° and 300° respectively, to real axis.

Fig. 7.3

Rule 6: Sometimes there are points in s-plane at which multiple roots of the characteristic equation occurs. These are called breakaway points and can be found dK =0. out by solving equation ds

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Consider a general characteristic equation as 1 + G (s) H (s) = 0. Assume that this equation has multiple roots (× times) at s = – y. Then this equation can be rewritten as 1 + G (s) H (s) = (s + y) × P (s); where P (s) is the remaining polynomial not having factor (s + y). Differentiating this equation with respect to s,

d ÈG (s) H (s)˘˚ = x (s + y ) x – 1 P (s) + (s + y ) x P ¢ (s) ds Î = (s + y)x – 1 [xP (s) + (s + y) P ¢(s)]

d P (s) ds d (7.16) Now, at s = – y, [G (s ) H (s )] = 0 ds Equation (7.16) has a root atleast order one at the same location as the multiple root of original characteristic equation. Thus, equation (7.16) gives the breakaway points. Now, the characteristic equation can be rewritten in pole-zero form as: where

P ¢(s) =

1 + G (s) H (s) = 1 + K

fi

K= –

C (s) =0 B (s)

(7.17)

B (s) C (s)

Differentiating K with respect to s,

dK È B (s) C ¢ (s) - C (s) B¢ (s) ˘ ˙=0 =Í 2 ds Í ˙˚ C ( s ) È ˘ Î ˚ Î

(7.18)

Again, differentiating equation (7.17) with respect to s, È B (s) C ¢ (s) - C (s) B¢ (s) ˘ d ˙=0 [G ( s ) H ( s ) = K Í 2 ds ÍÎ ˙˚ ÈÎ B (s)˘˚

fi

B (s) C¢ (s) – C (s) B¢ (s) = 0 Putting equation (7.19) in equation (7.18),

dK =0 ds

(7.19) (7.20)

Thus, breakaway points of the original characteristic equation are determined from the solution of either equation (7.20) or equation (7.16). Example 7.3 Find the breakaway point for the transfer function of Example 7.1, for drawing root locus.

Root Locus Technique 153

Solution: Step 1: As the two root locus branches start from 0 and – 1 and move towards each other, so a breakaway point must exist between 0 and – 1. Step 2: Characteristic equation is s (s + 1) (s + 3) + K = 0 K = – s (s + 1) (s + 3)

fi fi

K = – s3 – 4 s2 – 3 s dK Differentiating with respect to s, = – 3 s2 – 8 s – 3 . ds

Step 3: Breakaway points are the solution of equation fi

– 3 s2 – 8 s – 3 = 0 fi s = – 2.22, – 0.452.

dK =0 ds

Step 4: As breakaway point lies between 0 and – 1, so it is at – 0.452. Step 5: Breakaway point is shown in Fig. 7.4.

Fig. 7.4

Rule 7: The root locus ranches if intersect the imaginary axis, then the point of intersection can be determined using Routh’s Hurwitz criterion. Example 7.4 Find the point of intersection of root locus branches with imaginary axis for the transfer function of Example 7.1, for drawing root locus. Solution: Step 1:

G (s) H (s) =

K s (s + 1) (s + 3)

Characteristic equation is K + s (s + 1) (s + 3) = 0 fi s3 + 4 s2 + 3 s + K = 0

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Step 2: Routh’s array becomes, 1 s3 4 s2 12 – K s1 4 s0 K

3 K 0

Step 3: For the system to be stable, all the terms in the first column of Routh’s array 12 – K > 0. must be positive. Thus, K > 0 and 4 12 – K Step 4: For marginally stable system, = 0 fi K = 12 4 For this value of K, auxiliary polynomial A (s) = 4 s2 + 12 = 0 fi

s = ± j 1.732

Step 5: Root locus branches will intersect the imaginary axis at point ± j 1.732 as shown in Fig. 7.5.

Fig. 7.5

Rule 8: If complex poles and zeros exist in the control system, then angle of departure of branches from complex poles and angle of arrival of branches at complex zeros must be found out. The angle of departure from complex poles is given by fp = ± 180° (2q + 1) + f; q = 0, 1, 2, º

Root Locus Technique 155

q has different values for repeated complex poles and f is net angle contribution of all other poles and zeros at that complex pole. fz = ± 180° (2q + 1) – f; q = 0, 1, 2, º q has different values for repeated complex zeros and f is net angle contribution of all other poles and zeros at that complex zero. As the angle of departure from a real pole and angle of arrival at a real zero is always 180°, so these angles are needed to be found out only for complex poles and complex zeros respectively. Consider a pole-zero configuration as shown in Fig. 7.6. There are two real poles and one real zero. Also, there is one pair of complex poles and one pair of complex zeros in the system. To find the angle of departure at pole P1, fp 1 = + 180° (2q + 1) + f; q = 0 = 180° + f Now, f = q4 + q5 + q6 – q1 – q2 – q3

Fig. 7.6

Summary of Rules (i) Root locus is symmetrical about real axis in s-plane. (ii) The number of root locus branches is equal to the number of open loop poles (n). Each branch starts from one open loop pole and terminates on one open loop zero. The remaining (n – m) branches (m is the number of zeros) terminate on •. (iii) Any point on real axis lies on root locus iff the sum of open loop poles and zeros on real axis to the right hand side of that point is odd. So, real axis gets divided into alternate on-locus and not-on-locus segments. (iv) The angle of asymptotes (direction along which (n – m) branches tend towards •) is given by fa =

± 180∞ (2q + 1) ; q = 0, 1, 2, º, (n – m – 1) n-m

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(v) The point of intersection of asymptotes with real axis is centroid and is given by

sa =

 Real part of poles –  Real part of zeros n–m

(vi) The breakaway point at which multiple roots exist is the solution of equation dK = 0. ds (vii) The point of intersection of root locus branches with imaginary axis, if applicable, is determined using Routh’s Hurwitz criterion. (viii) The angle of departure from a complex pole, if applicable, is given by

fp = ± 180° (2q + 1) + f. Similarly, the angle of arrival from a complex zero, if applicable, is given by fp = ± 180° (2q + 1) – f; q = 0, 1, 2, º, (n – m – 1) where f is net angle contribution of all other poles and zeros at that complex pole or complex zero. K (s + 4) Example 7.5 Draw the root locus for unity feedback system having G (s) = . s (s + 2) (s + 6) Solution: K (s + 4) Step 1: Rule 1: Open loop transfer function is G (s) H (s) = s (s + 2) (s + 6) Poles are at s = 0, – 2, – 6. So, n = 3 Zeros are at s = – 4. So, m = 1 As all the poles and zeros are real, root locus will be symmetrical about real axis.

Step 2: Rule 2: There will be three branches in root locus each starting from s = 0, – 2 and – 6. Out of these three branches, one will terminate at s = – 4 and two will terminate on •. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.7.

Fig. 7.7

Root Locus Technique 157

The dark portions of real axis are the on-locus segments. It is clear that the on-locus and not-on-locus segments alternate on real axis. Step 4: Rule 4: The angle of asymptotes is ± 180∞ (2q + 1) ; q = 0, 1 fa = n-m = 90°, 270°. Step 5: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros = (0 – 2 – 6) – (– 4) = – 2 n–m

3–1

dK = 0. ds Now, characteristic equation is s (s + 2) (s + 6) + K (s + 4) = 0

Step 6: Rule 6: Breakaway point is the solution of

fi fi

- (s 3 + 8 s 2 + 12 s ) dK (s + 4) ( - 3 s 2 + 16 s - 12) + (s 3 + 8 s 2 + 12 s ) = =0 fi K= ( s + 4) ds ( s + 4 )2 s = – 1.15

Step 7: It is clear that there will be no intersection of root locus branches with imaginary axis. Also, there is no need to find angle of departure and angle of arrival in absence of complex poles and zeros. So, complete root locus is shown in Fig. 7.7. One branch will start from s = – 6 and terminate at s = – 4. Two branches will start from s = 0 and s = – 2, meet at the breakaway point at s = – 1.15 and proceed towards infinity along 90° and 270°. K . Example 7.6 Draw the root locus for the system having G (s) H (s) = s (s + 2) (s + 4) Solution: K Step 1: Rule 1: Open loop transfer function is G (s) H (s) = s (s + 2) (s + 4) Poles are at s = 0, – 2, – 4. So, n = 3 There are no zeros. So, m = 0 As all the poles are real, root locus will be symmetrical about real axis. Step 2: Rule 2: There will be three branches in root locus each starting from s = 0, – 2 and – 4. All of these will terminate on •. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.8. The dark portions of real axis are the on-locus segments. Step 4: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ; q = 0, 1, 2 n-m = 60°, 180°, 300°.

fa =

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Fig. 7.8

Step 5: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros = (0 – 2 – 4) – (0) = – 2 n–m

3–0

dK = 0. ds Now, characteristic equation is s (s + 2) (s + 4) + K = 0 dK = – 3 s2 – 12 s – 8 = 0. fi K = – (s3 + 6 s2 + 8 s) fi ds fi s = – 0.85, – 3.15

Step 6: Rule 6: Breakaway point is the solution of

As only – 0.85 lies on root locus, so – 0.85 is breakaway point. Step 7: Rule 7: Intersection of root locus branches with imaginary axis is found as: Characteristic equation is s (s + 2) (s + 4) + K = 0 fi s3 + 6 s2 + 8 s + K = 0 s3 1 s2 6 s1 48 – K 6 s0 K

8 K 0

48 – K = 0 fi K = 48 6 For this value of K, auxiliary polynomial A (s) = 6 s2 + 48 = 0

For marginally stable system,

fi

s = ± j 2.8 So, two branches cut the imaginary axis at point ± j 2.8 while going towards •. The complete root locus is shown in Fig. 7.8.

Root Locus Technique 159

Example 7.7

Sketch the root locus for the system having G (s) =

and H (s) = 1.

K ( s + 2) s (s + 3) (s 2 + 2 s + 20)

Solution: Step 1: Rule 1: Open loop transfer function is G (s) H (s) =

K ( s + 2) s (s + 3) (s 2 + 2 s + 20)

Poles are at s = 0, – 3, – 1 ± j 4. So, n = 4 Zeros are at s = – 2. So, m = 1 As all the poles and zeros are either real or complex conjugate pairs, root locus will be symmetrical about real axis. Step 2: Rule 2: There will be four branches in root locus each starting from s = 0, – 3 and – 1 ± j 4. Out of these four branches, one will terminate at s = – 2 and the rest three will terminate on •. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.9. Im

j4.26 180°

–3

–2

60° Re

–1 300° –j4.26

Fig. 7.9

The dark portions of real axis are the on-locus segments. Step 4: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ; q = 0, 1, 2 n-m = 60°, 180°, 300°.

fa =

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Step 5: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros = (0 – 3 – 1 – 1) – (– 2) = – 1 n–m

4–1

Step 6: Rule 6: No breakaway point exists. Step 7: Rule 7: Intersection of root locus branches with imaginary axis is found as: Characteristic equation is s (s + 3) (s2 + 2 s + 20) + K = 0 = s4 + 5 s3 + 26 s2 + 60 s + Ks + K = 0 Routh’s array is s4 s3 s2 s1 s0

1 26 2K 5 (60 + K ) 70 – K 2K 5 – K 2 – 40 K + 4200 0 70 – K 2K

- K 2 - 40 K + 4200 For marginally stable system, = 0 fi K = 45.12 70 - K 70 - 45.12 2 s + 2 (45.12) = 0 For this value of K, auxiliary polynomial A (s) = 5 fi s = ± j 4.26 So, two branches cut the imaginary axis at point ± j 4.26 while going towards •. Step 8: Rule 8: As complex conjugate poles exist in the system, angle of departure for these poles are found out using Fig. 7.10.

Fig. 7.10

Root Locus Technique 161

fp1 = + 180° (2q + 1) + f; q = 0 = 180° + f Now, f = q4 – q1 – q2 – q3 Now, q2 = + 90° q1 = + 90° + tan–1 (1/4) = 104.04° q3 = + tan–1 (4/2) = 63.43° q4 = + tan– 1 (4/1) = 75.96° So, f = 181.51° Hence, fp 1 = 180° – 181.51° = – 1.51° So, angle of departure for (– 1 + j 4) = – 1.51° and angle of departure for (– 1 – j 4) = + 1.51° The complete root locus is shown in Fig. 7.9. Example 7.8

Draw the root locus for the open loop transfer function of a control K ( s + 2) system is given by 2 (s + 0.4 s + 0.4) Solution: K ( s + 2) Step 1: Rule 1: Open loop transfer function is G (s) H (s) = 2 (s + 0.4 s + 0.4) Poles are at s = – 0.2 ± j 0.6. So, n = 2 Zeros are at s = – 2. So, m = 1 As all the poles and zeros are either real or complex conjugate pairs, root locus will be symmetrical about real axis. Step 2: Rule 2: There will be two branches in root locus each starting from (– 0.2 + j 0.6) and (– 0.2 – j 0.6). Out of these two branches, one will terminate at s = – 2 and one will terminate on •. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.11.

Fig. 7.11

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The dark portions of real axis are the on-locus segments. Step 4: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ;q=0 n-m = 180°

fa = Step 5: Rule 5: Centroid

 Real part of poles –  Real part of zeros = (– 0.2 – 0.2) – (– 2) = 1.6

sa =

n–m

2–1

dK = 0. ds Now, characteristic equation is s2 + 0.4 s + 0.4 + K (s + 2) = 0

Step 6: Rule 6: Breakaway point is the solution of

fi

K= -

(s 2 + 0.4 s + 0.4) dK - (s + 2) (2s + 0.4) + (s 2 + 0.4 s + 0.4) = =0 fi ds s+2 (s + 2) 2 s = – 3.89, – 0.1

fi

As only – 3.89 lies on root locus, so – 3.89 is breakaway point. Step 7: Rule 7: No branch will cut the imaginary axis. Step 8: Rule 8: As complex conjugate poles exist in the system, angle of departure for these poles are found out using Fig. 7.12.

Fig. 7.12

fp 1 = + 180° (2q + 1) + f; q = 0 = 180° + f Now,

f = q2 – q1

Now,

q1 = + 90°

Root Locus Technique 163

q2 = + tan– 1 (0.4/1.8) = 12.53° So,

fp 1 = 180° + 12.53° – 90° = 102.53°

So, angle of departure for (– 0.2 + j 0.6) = + 102.53° and angle of departure for (– 0.2 – j 0.6) = – 102.53° The complete root locus is shown in Fig 7.11. K Example 7.9 Draw the root locus of the system having G (s) H (s) = s (s + 4) (s 2 + 4 s + 20) Solution: K Step 1: Rule 1: Open loop transfer function is G (s) H (s) = s (s + 4) (s 2 + 4 s + 20) Poles are at s = 0, – 4, – 2 ± j 4. So, n = 4 There are no zeros. So, m = 0 As all the poles and zeros are either real or complex conjugate pairs, root locus will be symmetrical about real axis. Step 2: Rule 2: There will be four branches in root locus each starting from each pole. All of these will terminate on •. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.13. The dark portions of real axis are the on-locus segments.

Fig. 7.13

Step 4: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ; q = 0, 1, 2, 3 n-m = 45°, 135°, 225°, 315°

fa = Step 5: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros = (0 – 4 – 2 – 2) – (0) = – 2 n–m

4–0

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dK = 0. ds Now, characteristic equation is s (s + 4) (s2 + 4 s + 20) + K = 0

Step 6: Rule 6: Breakaway point is the solution of

4 3 2 K = – s - 8 s - 36 s - 80 s fi

fi

s = – 2, – 2 ± j 2.45

fi

dK = 4 s 3 + 24 s 2 + 72 s + 80 = 0 ds

As all the three breakaway points lie on the root locus, two branches will start from – 4 and 0, breakaway at – 2, then one will proceed towards (– 2 + j 4) and other towards (– 2 – j 4). These will again break away at two points. First branch will break away with (– 2 + j 4) at (– 2 + j 2.45) and second with (– 2 – j 4) at (– 2 – j 2.45) and then all will proceed towards •. Step 7: Rule 7: Intersection of root locus branches with imaginary axis is found as: Characteristic equation is s4 + 8 s3 + 36 s2 + 80 s + K = 0 Routh’s array becomes s4 3

s s2

s1 s0

1 8 26 2080 – 8 K 26 K

36 K 80 K 0

2080 – 8 K = 0 fi K = 260 26 For this value of K, auxiliary polynomial A(s) = 26 s2 + 260 = 0 fi s = ± j 3.16 So, two branches cut the imaginary axis at point ± j 3.16 while going towards •.

For marginally stable system,

Step 8: Rule 8: Branches will have to depart from complex poles (– 2 ± j 4) at ± 270° respectively so as to intersect with the other two branches at breakaway points. So, angle of departure is to be found out for complex breakaway points instead of complex poles. It is shown in Fig. 7.14. fb = +180° (2q + 1) + f; q = 0 = 180° + f Now,

f = – q2 – q1 – q3 – q4

Now,

q1 = + 270°; q4 = + 90° q2 = + 90° + tan–1 (2/2.45) = + 39.23° q3 = + tan–1 (2.45/2) = + 50.77°

So,

fb = – 270°

Root Locus Technique 165

Fig. 7.14

So, angle of departure for (–2 + j 2.45) = – 270° and angle of departure for (– 2 – j 2.45) = + 270° The complete root locus is shown in Fig. 7.13. Example 7.10

Sketch the root locus of the system having G (s) H (s) =

Solution:

K (s + 2) (s + 3) (s + 5) (s 2 + 2 s + 2)

K (s + 2) Step 1: Rule 1: Open loop transfer function is G(s) H (s) = (s + 3) (s + 5) (s 2 + 2 s + 2) Poles are at s = – 3, – 5, –1 ± j. So, n = 4 Zero is at s = – 2. So, m = 1 As all the poles and zeros are either real or complex conjugate pairs, root locus will be symmetrical about real axis. Step 2: Rule 2: There will be four branches in root locus each starting from each pole. Out of these four branches, one will terminate on one zero and rest will terminate on •. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.15. The dark portions of real axis are the on-locus segments. Step 4: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ; q = 0, 1, 2 n-m = 60°, 180°, 300°

fa =

Step 5: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros

n–m ( -3 - 5 - 1 - 1) - ( -2) = -2 = 4-1

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Fig. 7.15

Step 6: Rule 6: Step 7: Rule 7: Characteristic fi Routh’s array

Breakaway point does not exist. Intersection of root locus branches with imaginary axis is found as: equation is (s + 3) (s + 5) (s2 + 2 s + 2) + K (s + 2) = 0 s4 + 10 s3 + 33 s2 + (46 + K) s + (30 + 2 K) = 0 becomes 33 (30 + 2 K ) s4 1 (46 + K ) s 3 10 284 – K (30 + 2 K ) s2 10 – K 2 + 38 K + 10064 s1 0 284 – K s0 (30 + 2 K )

- K 2 + 38 K + 10064 = 0 fi K = 121.2 284 - K 284 - 121.2 2 For this value of K, auxiliary polynomial A (s) = s + (30 + 2 (121.2)) = 0 10 fi s = ± j 4.09 For marginally stable system,

So, two branches cut the imaginary axis at point ± j 4.09 while going towards •. Step 8: Rule 8: As complex conjugate poles exist in the system, angle of departure for these poles are found out using Fig. 7.16.

Now, Now,

fp 1 = = f= q2 = q1 =

+ 180° (2q + 1) + f; q = 0 180° + f – q4 + q1 – q2 – q3 + 90° + tan–1 (1) = + 45°

Root Locus Technique 167

Fig. 7.16

q3 = + tan–1 (1/2) = + 26.57° q4 = + tan–1 (1/4) = + 14.04° Hence, fp 1 = + 94.39° So, angle of departure for (– 1 + j 1) = + 94.39° and angle of departure for (– 1 – j 1) = – 94.39° The complete root locus is shown in Fig. 7.15.

7.4

DETERMINATION OF K FOR A SPECIFIED DAMPING FACTOR

The following steps are followed to determine the value of system gain K corresponding to a specified damping factor d using root locus technique: (i) Draw root locus of the system for the given open loop transfer function. (ii) Draw a damping line from origin at an angle q with negative real axis in clockwise direction where cos q = d. (iii) Locate the point of intersection of this damping line with the root locus. Let it be point S. (iv) Find the coordinates of S using graph scale. (v) Find the value of K corresponding to this S using magnitude condition. Example 7.11 Using the control system of Example 7.6, determine K for damping factor 0.5 using root locus technique. Solution: Step 1: Root locus for the system having G (s) H (s) =

K is again shown s (s + 2) (s + 4)

in Fig. 7.17. Step 2: cos q = 0.5 fi q = 60°, S is (– 0.75 + j 0.8) approximately.

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60° S –4

–2

Re

Fig. 7.17

Step 3: To find K, magnitude condition for transfer function is used.

G (s) H (s) s = ( - 0.75 + j 0.8) = 1 fi

K =1 s (s + 2) (s + 4) s = ( - 0.75 + j 0.8)

fi

K =1 - 0.75 + j 0.8 1.25 + j 0.8 3.25 + j 0.8

fi

K = 1 fi K = 5.40 1.09 ×1.48 × 3.35

7.5

CANCELLATION OF POLES AND ZEROS

If in the open loop transfer function G (s) H (s) of the system, there is one common factor in both numerator and denominator i.e., pole and zero at the same location, then these factors cancel each other and root locus plotted will not include these poles and zeros. In such cases, where cancellation of poles and zeros occur, the closed loop poles of the system are obtained from root locus plus the cancelled poles of the system. Or, the root locus calculations can be modified by adding that open loop pole to the G (s) H (s) of the system, to get the actual root locus. K and Example 7.12 Sketch the root locus for system having G (s) = (s + 2) (s + 2) H (s) = (s 2 + 4 s + 5) Solution: Step 1:

G (s) H (s) =

K (s + 2) K = 2 2 (s + 2) (s + 4 s + 5) (s + 4 s + 5)

Root Locus Technique 169

As one open loop pole cancels with one open loop zero, so to do the calculations for the rest of the root locus, G (s) H (s) is taken as K G (s) H (s) = (s + 2) (s 2 + 4 s + 5) Step 2: Rule 1: Poles are at s = – 2, – 2 ± j. So, n = 3 and there are no zeros. So, m = 0. As all the poles are either real or complex conjugate pairs, root locus will be symmetrical about real axis. Step 3: Rule 2: There will be three branches in root locus each starting from each pole and all will terminate on •. Step 4: Rule 3: The pole-zero location is shown in Fig. 7.18. The dark portions of real axis are the on-locus segments. Step 5: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ; q = 0, 1, 2 n-m = 60°, 180°, 300°

fa =

Step 6: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros

( -2 - 2 - 2) - (0) = -2 = 3-0

n–m

Fig. 7.18

Step 7: Rule 6: Breakaway point does not exist. Step 8: Rule 7: Intersection of root locus branches with imaginary axis is found as: Characteristic equation is (s + 2) (s2 + 4 s + 5) + K = 0 fi s3 + 6 s2 + 13 s + 10 + K = 0

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Routh’s array becomes s3 s2 s1 s0

1 13 6 (10 + K ) 68 – K 0 6 (10 + K )

68 – K = 0 fi K = 68 6 For this value of K, auxiliary polynomial A (s) = 6 s2 + 78 = 0 fi s = ± j 3.6 So, two branches cut the imaginary axis at point ± j 3.6 while going towards •.

For marginally stable system,

Step 9: Rule 8: As complex conjugate poles exist in the system, angle of departure from these poles are found out using Fig. 7.19.

Fig. 7.19

fp 1 = + 180° (2q + 1) + f; q = 0 = 180° + f Now,

f = – q1 – q2

Now,

q1 = + 90°; q2 = + 90° f = – 90° – 90° = 180°

Hence,

fp 1 = 0°

So, angle of departure for (– 2 + j 1) = 0° and angle of departure for (– 2 – j 1) = 0° The complete root locus is shown in Fig. 7.18.

Root Locus Technique 171

7.6

ROOT LOCUS WHEN m > n

When in a control system, the order of numerator (m) is greater than that in denominator (n), then the number of branches in root locus is equal to m. Out of these m branches, n will start from each pole and rest (n – m) will start from •. All the branches will terminate on each zero. Example 7.13 G (s) H (s) = Solution:

Draw the root locus for the system having open loop transfer function

K (s + 2) (s 2 + 2 s + 2) (s + 1)

Step 1: Rule 1: Pole is at s = – 1. So, n = 1 Zeros are at s = – 2, – 1 ± j. So, m = 3 As all the poles and zeros are either real or complex conjugate pairs, root locus will be symmetrical about real axis. Step 2: Rule 2: There will be three branches in root locus. One will start from one pole and rest two will start from •. All the three branches will terminate on each zero. Step 3: Rule 3: The pole-zero location is shown in Fig. 7.20. The dark portions of real axis are the on-locus segments.

Fig. 7.20

Step 4: Rule 4: The angle of asymptotes is

± 180∞ (2q + 1) ; q = 0, 1 m-n = 90°, 270°

fa =

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Step 5: Rule 5: Centroid sa =

 Real part of poles –  Real part of zeros

=

n–m ( - 1) - ( - 2 - 1 - 1) = - 1.5 1-3

Step 6: Rule 6: Breakaway point does not exist. Step 7: Rule 7: No intersection of root locus branches with imaginary axis. Step 8: Rule 8: As complex conjugate zeros exist in the system, angle of arrival at these zeros is found out using Fig. 7.21.

Fig. 7.21

fz 1 = + 180° (2q + 1) – f; q = 0 = 180° – f Now, f = – q1 + q2 + q3 Now, q1 = + 90°; q2 = + 90° q3 = + tan–1 (1) = + 45° Hence, fz 1 = + 135° So, angle of arrival for (– 1 + j 1) = + 135° and angle of arrival for (– 1 – j 1) = – 135° The complete root locus is shown in Fig. 7.20.

7.7

ROOT CONTOURS

As already discussed, root locus is a graphical plot of locus of roots of the control system, when some system parameter is varied from zero to infinity. Generally, this parameter is system gain K. Sometimes, some other parameter of the system can be varied in addition to K and corresponding root locus plots are called root contours.

Root Locus Technique 173

Consider a feedback control system having open loop transfer function as K G (s) H (s) = s (s + 1) (s + a) In this system, system gain K and pole s = – a are both variable parameters. Characteristic equation becomes, 1+ fi fi

K =0 s (s + 1) (s + a)

s2 (s + 1) + as (s + 1) + K = 0 as (s + 1) 1+ 2 =0 s (s + 1) + K

(7.21)

(7.22)

Equation (7.22) is the rearranged form of equation (7.21), so that root locus can be plotted in terms of a, by keeping K constant. This is done by firstly drawing root locus by varying K and keeping a equal to zero. The characteristic equation when a = 0 becomes, K =0 1+ 2 s (s + 1) So, poles are at s = 0, 0, – 1. n = 3 There are no zeros. m = 0 ± 180∞ (2q + 1) ; q = 0, 1, 2 fa = n-m = 60°, 180°, 300°

Fig. 7.22

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Centroid  Real part of poles –  Real part of zeros = (0 + 0 – 1) – (0) = – 0.333 sa = n–m 3–1 The root locus is shown in Fig. 7.22. This root locus starts from initial value of K = 0, then K goes on increasing upto •. Accordingly, points are located on this root locus as K1, K2, K3, º. Now, keep the value of K constant i.e., K = K1 and vary a from 0 to • in equation (7.22) to get one root locus with respect to a. This is shown in Fig. 7.23. Now, different root loci are obtained for different values of K like K1, K2, K3, • and by varying a from 0 to •. These set of root loci are root contours. These are shown in Fig. 7.23. Im

K = K3 K = K2 K = K1

K=0 K = K3 K = K2 K = K1

K=0

–1

Fig. 7.23

Exercise 1. Sketch the root locus for the open loop transfer function

G (s) H (s) =

K . (s + 1) (s + 2) (s + 4) (s + 5)

Re

Root Locus Technique 175

2. The open loop transfer function of a control system is given by G (s) H (s) = K . Find the breakaway points, angle of departure from complex s (s + 6) (s 2 + 4 s + 13) poles and draw root locus. Also, determine the conditions for stability. K (s 2 + 2 s + 2) . 3. Plot the root locus for the system having G (s) H (s) = s (s 2 + 4 s + 8) 4. Plot the root locus for the control system shown in Fig. 7.24.

Fig. 7.24

5. Using root locus technique, find gain K for damping factor = 0.341 for the unity K . feedback control system having G (s) = s (s + 2) (s + 3) K (s 2 - 2 s + 5) . 6. The open loop transfer function of a control system is G (s) H (s) = (s 2 + 1.5 s - 1) Draw root locus and find gain K for damping ratio = 0.5. K (0.2 s 2 + 0.3 s + 1) 7. The transfer function of a control system is G (s) H (s) = . s (s 2 + 0.2 s - 1) Construct root locus and find out critical value of K.

Objective Type Questions 1. A system has four poles and two zeros. The number of branches of root locus terminating on • are (a) 2 (b) 1 (c) 4 (d) 0 2. The branches of root locus start from (a) open loop zeros (b) closed loop zeros (c) open loop poles (d) closed loop poles 3. The branches of root locus can terminate on (a) open loop zeros (b) closed loop zeros (c) open loop poles (d) closed loop poles

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K 4. For a transfer function , the number of branches terminating on • 2 s ( s + 6 s + 9) are (a) 1 (b) 2 (c) 3 (d) 4 K 5. The centre of asymptotes for transfer function is 2 s (s + 4 s + 13) (a) –1 (b) – 1.33 (c) – 2 (d) – 2.5 K 6. The transfer function of a system is . The number of asymptotes s (s + 2) (s + 4) in root locus is

(a) 1 (b) 2 (c) 3 (d) 4 7. Breakaway points in root locus occur on (a) real axis (b) imaginary axis (c) multiple roots of characteristic equation (d) either (a) or (b) K 8. The open loop transfer function is . Its centroid in the root locus s ( s + 1) ( s + 2) is at (a) 0 (b) 2 (c) – 1 (d) – 2 9. If the root locus has two asymptotes, the system may have (a) 2 poles (b) 4 poles and 2 zeros (c) 3 poles and 1 zero (d) all of these 10. The equation used for determining breakaway point is dK =0 (b) Ú Kds = 0 (a) ds dG (s ) (c) =0 (d) |G (s) H (s)| = 0 ds K 11. For a system having transfer function G (s) = , breakaway s (s + 1) (s 2 + 4 s + 5) point occur at (a) – 2 (b) – 0.4 (c) – 4 (d) – 2.4 12. Intersection of root locus branches with imaginary axis is found out using (a) Bode plot (b) Nyquist criterion (c) Nicholas charts (d) Routh’s Hurwitz criterion

Root Locus Technique 177

13. Angle of arrival in root locus is calculated for (a) complex poles (b) complex zeros (c) real poles (d) real zeros 14. Angle of departure in root locus is calculated for (a) complex poles (b) complex zeros (c) real poles (d) real zeros K 15. The angle of asymptotes for transfer function are 2 s (s + 4 s + 8)

(a) 0°, 60°, 120° (c) 60°, 180°, 300°

(b) 0°, 180°, 270° (d) 90°, 270°, 360°

Answers 1. (a) 6. (c) 11. (b)

2. (c) 7. (c) 12. (d)

3. (a) 8. (c) 13. (b)

4. (c) 9. (d) 14. (a)

5. (b) 10. (a) 15. (c)

MATLAB Programs P1. Sketch the root locus for the open loop transfer function

Program: n=[0 1]; d=[1 6 5 10 0]; T=tf(n,d); rlocus(T)

Execution:

% % % %

Numerator of transfer function Denominator of transfer function Transfer function plots the root locus

K . (s + 6 s + 5 s 2 + 10 s ) 4

3

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P2. Draw the root locus for the open loop transfer function

K (s + 1) . s (s 2 + 2 s + 2)

Program: n=[1 1]; d=[1 2 2 0]; T=tf(n,d); rlocus(T)

% Numerator of transfer function % Denominator of transfer function % Transfer function % plots the root locus

Execution:

P3. Draw the root locus for the open loop transfer function

Program: n=[0 1]; d=[1 5 12 8 0]; T=tf(n,d); rlocus(T)

Execution:

% % % %

Numerator of transfer function Denominator of transfer function Transfer function plots the root locus

K . s (s + 1)(s 2 + 4 s + 8)

8

Frequency Response Analysis

CHAPTER

In previous chapters, time response analysis of a system with different standard test signals has been studied. When the same system is supplied with sinusoidal input, then the analysis is referred as frequency response analysis. Consider a linear time-invariant system with sinusoidal input r (t) as: r (t) = P sin wt In steady state, system output is also sinusoidal and can be written as: c (t) = Q sin (wt + F) The analysis of magnitude and phase angle of sinusoidal input r (t) and steady state output c (t) is known as frequency response analysis. The frequency response can be easily evaluated from the sinusoidal transfer function which can be obtained by replacing s by j w in transfer function T (s). The transfer function T ( jw) thus obtained, is a complex function having both magnitude and phase angle. These characteristics are easily studied by using graphical plots. The various plots studied here are: (i) Bode Plot (ii) Polar Plot (iii) Nyquist Plot

8.1 CORRELATION BETWEEN TIME RESPONSE AND FREQUENCY RESPONSE OF A SECOND ORDER SYSTEM Consider the transfer function of a second order system in time domain: C (s) w n2 = 2 T (s) = R (s ) s + 2 dw n s + w n2 where d = damping factor wn = undamped natural frequency of oscillations For sinusoidal transfer function, replace s by jw, C ( jω ) ω n2 = R ( jω ) ( jω )2 + 2 δω n ( jω ) + ω n2 Dividing by wn2, T ( jw) =

1 2

Ê wˆ Ê wˆ ÁË j w ˜¯ + 2 d ÁË j w ˜¯ + 1 n n

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w = u, wn

Putting

T ( jw) = Now,

1 1 = 2 - u + 2 δ ju + 1 (1 - u ) + j 2 δu 2

M = |T( jw)| =

1 (1 - u ) + (2 δu)2

-1 F = –T ( jw) = − tan

And

2 2

2 du (1 - u2 )

(8.1) (8.2)

The frequency where M has a peak value is known as resonant frequency (wr). To find wr,

dM du

=0 u = ur

At this point, slope of the M-curve i.e., M versus w curve is zero as shown in Fig. 8.1(a). From equation (8.1), -

2 2 1 ÈÎ 2 (1 - ur ) ( - 2 ur ) + 8 d ur ˘˚ =0 3 2 2 2 2 2 ÈÎ(1 - ur ) + (2 δur ) ˘˚

fi

- 4 (1 - ur2 ) + 8 δ 2 = 0

fi

ur2 = 1 - 2 δ 2

fi

ur = 1 - 2 d 2

fi fi

wr = 1 - 2 δ2 ωn wr = wn = 1 – 2 d 2

(8.3)

This resonant frequency is indicative of u i.e., normalized frequency and of settling time. The maximum value of magnitude M as in equation (8.1) is called resonant peak (Mr). 1 Mr = [1 - (1 - 2 δ 2 )]2 + (2 δ 1 - 2 δ 2 )2 1 = 4 4 δ + 4 δ 2 (1 − 2 δ 2 ) 1 = 4 4 δ + 4 δ2 - 8 δ4 1 = 2 4 δ (1 - δ 2 ) 1 Mr = (8.4) 2 δ 1 - δ2

Frequency Response Analysis 181

This resonant peak is indicative of damping factor. The phase angle at resonant frequency can be obtained from equation (8.2), by putting u as ur , fr = − tan -1 -1 fr = − tan

2 d 1 - 2 d2 1 - 1 + 2 d2 1 - 2 d2 d2

(8.5)

Fig. 8.1

The graph representing curve of M versus w is shown in Fig. 8.1, using equations (8.1), (8.3) and (8.4). As d approaches zero, wr approaches wn and Mr approaches •. For 0 < d < 0.707, wr is always less than wn and Mr is greater than 1. For d ≥ 0.707, Mr = 1 i.e., peak of the curve vanishes. It can be concluded from above discussion that for a second order system, Mr is indicative of d for 0 < d ≤ 0.707 and wr is indicative of its natural frequency for given d and hence indicative of its speed of response or setting time i.e., ts = 4/(dwn). So, Mr and wr of frequency response can be used as performance indices of second order system. The frequency at which M has value of 0.707 is called cut off frequency (wc). Frequencies above wc are greatly attenuated from passing through a system. For closed loop systems, the range of frequencies over which M is equal to or greater than 0.707 is known as bandwidth (wb). This is shown in Fig. 8.2. Generally, wb of control system indicates the noise filtering characteristics of the system. Now, consider the step response of second order system. Expressions for damped frequency wd and peak overshoot Mp for 0 ≤ d ≤ 1 are wd = w n

1 – d2

(8.6)

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M Mr wb = wc – wr 1 0.707

wb wr

wc w

Fig. 8.2

Mp = e

– pd / 1 – d 2

(8.7)

Peak of curve vanishes when Mp = 0 i.e., when d = 1. The comparison of Mp and Mr is shown in Fig. 8.3(a) for various value of d. Figure 8.3(a) shows that for 0 < d ≤ 0.707, the two performance indices i.e., Mp and Mr are co-related, as both are functions of d. It means that, a system with a given value of Mr of frequency response must have a corresponding value of Mp if subjected to step input. For d > 0.707, step response oscillations are well damped. Similarly, equations (8.3) and (8.4) show the comparison between wr and wd. This is shown in Fig. 8.3(b). The ratio of these two frequencies is wr/wd = of d.

8.2

1 - 2 δ2 is function 1 - δ2

FREQUENCY RESPONSE SPECIFICATIONS

Frequency response specifications are performance indices of a system. Like in time response analysis, these are also defined for frequency response analysis. These are described below: (i) Resonant peak (Mr ): It is the maximum value of magnitude of closed loop 1 frequency response. It is given by: . Larger the value of Mr , more is 2 d 1 – d2 the overshoot of the system. (ii) Resonant frequency ( r): It is the frequency at which resonant peak occurs in closed loop frequency response. It is given by:

Frequency Response Analysis 183

Fig. 8.3

(iii) (iv) (v) (vi)

wr = ω n 1 - 2 δ 2 Bandwidth ( b): It is the range of frequencies over which M is greater than or equal to 0.707 or over which system responds satisfactorily. Gain crossover frequency ( gc): It is the frequency at which magnitude of closed loop frequency response is unity i.e. |G ( jw) H ( jw)| = 1. Phase crossover frequency ( pc): It is the frequency at which phase angle of closed loop frequency response is – 180°. i.e., – G ( jw) H ( jw) = – 180°. Gain Margin (GM): It is the margin by which the system gain can be increased to drive it to the verge of instability. It is reciprocal of magnitude of G ( jw) H ( jw) at wpc i.e. 1 GM = 20 log10 G ( jw ) H ( jw ) w = w pc

(vii) Phase Margin (PM): It is defined as the amount of additional phase lag at the gain cross-over frequency, required to bring the system to the verge of instability. Mathematically, PM = –G ( jw) H ( jw)|w = wgc + 180°

8.3

BODE PLOT

The frequency response of a closed loop system can be obtained by plotting its magnitude M and phase angle F against frequency. These are called Bode plots. So, Bode plots consist of two graphs: (a) Magnitude M expressed in logarithmic value 20 log10 |G ( jw) H ( jw)|against logarithmic frequency log10 (w).

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(b) Phase angle F in degrees against logarithmic frequency log10 (w). The standard procedure is to draw M i.e., 20 log10 |G ( jw) H ( jw)|and F(w) on linear scale and frequency w on log scale. Unit of M is decibel (db) i.e., on y – axis → linear scale for M in db and F in degrees. on x – axis → log scale for frequency. The range of frequency or frequency ratios are normally expressed in two units i.e., octave and decade. The range of frequencies w2 = 2 w1 or w2 = w1/2 is called octave and the range of frequencies w2 = 10 w1 or w2 = w1/10 is called decade.

8.3.1 Bode Plot of a General Transfer Function Consider a general transfer function, K (1 + sTa ) (1 + sTb ) º G ( s) H ( s) = (s )N (1 + sT1 ) (1 + sT2 ) … Frequency response transfer function is obtained by replacing s by jw, K (1 + j ωTa ) (1 + j ω Tb ) … fi |G ( jw) H ( jw)| = ( j ω )N (1 + j ω T1 ) (1 + j ω T2 ) … There are four basic factors in this transfer function: (i) Gain K (ii) Poles and zeros at origin i.e., ( jw)± N (iii) Real poles and zeros i.e., (1 + j wT )± N (iv) Quadratic or complex poles and zeros i.e., (s2 + 2dwns + wn2) Now, all these factors are considered one by one to draw Bode plot: (i) Gain K Here G (s) H (s) = K Replacing s by jw, G ( jw) H ( jw) = K + j 0 fi

|G ( jw) H ( jw)| = K Now, M = 20 log10 |G ( jw) H ( jw)| = 20 log K db.

As M is independent of w, so magnitude plot will be a straight line parallel to x-axis. If K > 1, then this line is above 0 db. For K < 1, this line is below 0 db. At K = 1, line is at 0 db. This is shown in Fig. 8.4. Now, phase angle

0 F = tan–1 ÊÁ ˆ˜ = 0; if K is positive Ë K¯

= – 180°; if K is negative. (ii)

Poles and zeros at origin i.e., ( j ) ± N Let us start with one pole at origin i.e., ( jw)– 1 (N = 1) 1 Thus, G (s) H (s) = jw

Frequency Response Analysis 185

Fig. 8.4

Fig. 8.5

Replacing s by j w, G ( jw) H ( jw) = fi

1 jw

M = 20 log

(8.8) 1 1 = 20 log = – 20 log w db jw w

This is similar to straight line equation y = mx, where m is the slope of the line. So, M = – 20 log w db equation will be a straight line with slope – 20 db at w = 1. Now, 10 times change in frequency range is 1 decade. Hence, slope of the plot for N = 1 is – 20 db/decade (or – 6 db/octave). In general, for N poles at origin, slope of the straight line becomes –20*N db/decade. This is shown in Fig. 8.6. Procedure to draw straight line with given slope: Suppose at w = w1, M = a is known i.e., coordinate X1 is known as (w1, a). Also, slope of the curve is known, as shown in Fig. 8.7. Now, procedure to obtain point X2 is as follows: w2 = 10 w1 and b = a + slope of the line. Thus, (w2, b) becomes the coordinate of point X2. Then, join X1 – X2.

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Fig. 8.6

Fig. 8.7

The phase of the equation (8.8) is given by: –1 Ê w ˆ F = – tan Á ˜ = – 90° (1 pole) Ë 0¯ = – 180° (2 pole) and so on. It is shown in Fig. 8.8.

Fig. 8.8

Frequency Response Analysis 187

If it is zero at origin, then G ( jw) H ( jw) = ( jw)1 (N = 1) Thus, M = 20 log |jw| = 20 log w db.

Fig. 8.9

Again, plot will be a straight line but with slope of + 20 db/decade. In general, for N zeros at origin, the slope is + 20* N db/decade. Also,

–1 Ê w ˆ F = + tan Á ˜ = + 90° (1 zero) Ë 0¯

= +180° (2 zero) and so on. This is shown in Fig. 8.9 (a) and (b). (iii)

Real poles or zeros i.e., (1 + j T) ± N

Let us start with real pole i.e., (1 + jwT )– 1 (N = 1) Here,

G ( jw) H ( jw) =

1 (1 + jωT )

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Now,

M = 20 log10 |G ( jw) H ( jw)| = 20 log10 = – 20 log10

1 (1 + jωT )

1 + w 2 T 2 = –10 log10 (1 + w2 T2) db.

This curve is plotted using method of approximations. Total frequencies are divided into two categories: low frequencies and high frequencies. 1 , M = – 10 log10 1 = 0 db. For low frequencies i.e., w > , M = – 10 log10 (w2 T 2) = –20 log10 (wT ) db. T 1 So, for high frequencies i.e., frequencies greater than frequency , curve is a straight T line of slope – 20 db/decade. So, M versus log w plot consists of two straight line 1 and other with slope of – 20 db/decade for • > w asymptotes, one at 0 db for 0 < w < T 1 1 at which the two asymptotes meet, is called corner frequency > . The frequency w = T T (wc.) or break frequency. The plot is called asymptotic magnitude plot as shown in Fig. 8.10.

Fig. 8.10

Similarly, for zeros, same procedure is followed with slope of + 20 db/decade. Now, this plot is approximated plot. Correct or exact plot can be obtained by applying error correction, which can be calculated as described below: 1 Error for low frequency range i.e. 0 < w < T (8.9) = – 10 log10 (1 + w2 T 2) – (– 10 log10 1)

Frequency Response Analysis 189

1 Now, correction at w = = – 10 log10 (1 + 1) + 10 log10 1 = – 10 log10 2 = – 3 db. T 1 = –10 log10 (1 + 1/4) + 10 log10 1 = – 10 log10 1 = – 1 db. and correction at w = 2T Similarly, corrections at other low frequencies can be obtained using equation (8.8). 1 Now, error for high frequency range i.e. • > w > T = – 10 log10 (w2 T 2) – (– 20 log10 wT ) (8.10) 1 = – 10 log10 (1 + 1) + 20 log10 1 = – 3 db. T 1 = – 10 log10 (1 + 4) + 20 log10 2 = – 1 db. w= 2T

Thus, correction at w = and correction at

Similarly, corrections at other high frequency can be obtained using equation (8.10). If we consider the case of zeros, then these corrections come out to be positive. So, it is concluded that accurate magnitude plot is obtained by correcting the asymptotic plot by 3 db for poles and zeros respectively at corner frequency and by 1 db for poles and zeros respectively at one octave above and below the corner frequency. A smooth curve is then drawn through these new points. It is also shown in Fig. 8.10. wT = – tan – 1 (wT ) Now, phase angle of G ( jw) H ( jw), F = – tan – 1 –1 1 , F = – 45° At w= T w=

1 , F = 0° 10T

10 , F = – 90° T This is shown in Fig. 8.11.

w=

Fig. 8.11

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(iv)

Complex poles and zeros: ω n2 = s2 + 2 δ ω n s + ω n2

1 Ê 2 δ s ˆ Ê s2 ˆ + 1+Á Ë ω n ˜¯ ÁË ω n2 ˜¯ Replacing s by jw, 1 1 = G ( jw) H ( jw) = 2 2 Ê jω ˆ Ê ωˆ Ê ωˆ 2δ 1+ 1+ j2δ Á ˜ - Á ˜ jω + Á ˜ ωn Ë ωn ¯ Ë ωn ¯ Ë ωn ¯

G (s) H (s) =

|G ( jw) H ( jw)| =

fi

=

Now,

(8.11)

1 2

È Ê ω ˆ 2 ˘ È Ê ω ˆ ˘2 Í1 - Á ˜ ˙ + Í 2 δ Á ˜ ˙ ÍÎ Ë ω n ¯ ˙˚ Î Ë ω n ¯ ˚ 1 2

È Ê ω ˆ2˘ Í1 - Á ˜ ˙ + 4 δ 2 ÍÎ Ë ω n ¯ ˙˚

M = 20 log10

ÈÊ ω ˆ ˘ ÍÁ ˜ ˙ ÎË ω n ¯ ˚

2

2 2 2˘ ÈÏ ¸ Í ÔÌ1 - Ê ω ˆ Ô˝ + 4 δ 2 Ê ω ˆ ˙ ÁË ω ˜¯ ÁË ω ˜¯ ˙ Í n n ˛Ô Î ÓÔ ˚

2 2 2˘ ÈÏ ¸Ô Ê ˆ ω Ô 2 Ê ω ˆ ˙ Í = – 10 Ì1 - Á ˜ ˝ + 4 δ Á ˜ Í Ë ω n ¯ ˛Ô Ë ωn ¯ ˙ Î ÓÔ ˚

-

-

1 2

1 2

Again, this curve is plotted using method of approximations. For low frequencies, w > wn and 4 d Á ˜ wgc. System is said to be unstable, when gain margin and phase margin are both negative. This happens, only when wpc < wgc. System is said to be marginally stable, when gain margin and phase margin are both zero. This happens, only when wpc = wgc. 50 . Example 8.1 A unity feedback system has transfer function G (s) = s (s + 2) (s + 5) Sketch Bode plot. Discuss stability of the system. Solution: Step 1: fi

G (s) H (s) = G ( jw) H ( jw) =

50 s (s + 2) (s + 5) 50 = jw ( jw + 2) ( jw + 5)

Here corner frequencies are 2 and 5 rad/s.

5 . jw ˆ Ê jw ˆ Ê jw Á 1 + ˜ Á 1 + ˜ Ë 2¯Ë 5¯

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5 Step 2: The first basic factor associated with G ( jw) H ( jw) is . jw Here, M = 20 log10 5 – 20 log10 w At w = 1, M = 14 db At w = 10 (decade system), M = 14 – 20 = – 6 db. These points are plotted on M versus log w graph as dotted line as shown in Fig. 8.16(a). This dotted line is made solid upto first corner frequency i.e., 2 rad/s. 1 Step 3: Next basic factor in G ( jw) H ( jw) is j wˆ Ê ÁË 1 + 2 ˜¯

From M versus w graph shown in Fig. 8.16(a), At w = 2, M = 8 db So, at w = 20 (decade system), M = previous M + Net slope (as discussed in section 8.3.1) = 8 + (– 20 – 20) = – 32 db. These points are joined by dotted line on M versus log w graph. This dotted line is made solid upto second corner frequency i.e. 5 rad/s. 1 Step 4: Next basic factor in G ( jw) H ( jw) is j wˆ Ê ÁË 1 + 5 ˜¯ From M versus log w graph shown in Fig. 8.16 (a), At w = 5, M = – 8 db So, at w = 50 (decade system), M = previous M + net slope (as discussed in section 8.3.1) = – 8 + (– 40 – 20) = – 68 db. These points are joined by dotted line on M versus log w graph. This line is made solid further. Step 5: To draw F versus log w graph, firstly phase angle equation is written: w w – tan– 1 F = – 90° – tan–1 2 5 The different values of F for different w are: w (rad/s) 1 2 5 10 F (°) –128 –157 –203 –232 This F versus log w graph is shown in Fig. 816(b). Step 6: From Bode plot, gain cross-over frequency, wgc = 3.1 rad/s and phase crossover frequency, wpc = 3.1 rad/s. (Answer) As wpc = wgc , so system is marginally stable.

Frequency Response Analysis 195 M(db) 20

(1, 14)

–20 db/decade (2, 8)

10

–40 db/decade

0

(10, –6)

(5, –8)

–10 –20

–60 db/decade

–30

(20, –32)

–40 –50 –60 –70 –80

(50, –68)

wgc = 3.1 rad/s 1

10

100 (a)

1000

w

100 (b)

1000

w

F(°) (1, –128) –140 (2, –157)

–160

–180

–200

(5, –203)

–220 wpc = 3.1 rad/s –240

1

(10, –232) 10

Fig. 8.16

(a), (b)

64 (s + 2) . s (s + 0.5) (s 2 + 3.2 s + 64) Sketch Bode plot. Discuss stability of the system using gain margin and phase margin.

Example 8.2 A unity feedback system has transfer function G (s) = Solution: Step 1:

G (s) H (s) =

64 (s + 2) s (s + 0.5) (s 2 + 3.2 s + 64)

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jw ˆ Ê 4Á1 + ˜ Ë 64 ( jw + 2) 2¯ = fi G ( jw) H (j w) = . 2 jw ( jw + 0.5) ( - w 2 + 3.2 jw + 64) jw ˆ È Ê w ˆ Ê Ê wˆ ˘ jw Á 1 + Í1 - Á ˜ + j 0.4 Á ˜ ˙ Ë 8 ¯ ˙˚ Ë 0.5 ˜¯ ÍÎ Ë 8 ¯

Here corner frequencies are 0.5, 2 and 8 rad/s.

4 Step 2: The first basic factor associated with G ( jw) H ( jw) is . jw Here, M = 20 log10 4 – 20 log10w At w = 1, M = 12 db At w = 10 (decade system), M = 12 – 20 = – 8 db. These points are plotted on M versus log w graph as dotted line as shown in Fig. 8.17(a). This dotted line is made solid upto first corner frequency i.e., 0.5 rad/s.

Step 3: Next basic factor in G ( jw) H ( jw) is

1 jw ˆ Ê ÁË 1 + 0.5 ˜¯

From M versus log w graph shown in Fig. 8.17(a), At

w = 0.5, M = 18 db

So, at w = 5 (decade system), M = previous M + net slope (as discussed in section 8.3.1) = 18 + (– 20 – 20) = – 22 db. These points are joined by dotted line on M versus log w graph. This dotted line is made solid upto second corner frequency i.e., 2 rad/s. jw ˆ Ê Step 4: Next basic factor in G ( jw) H ( jw) is Á 1 + ˜ Ë 2¯

From M versus log w graph shown in Fig. 8.17(a), At w = 2, M = – 6 db So, at w = 20 (decade system), M = previous M + Net slope (as discussed in section 8.3.1) = – 6 + (– 40 + 20) = – 26 db. These points are joined by dotted line on M versus log w graph. This dotted line is made solid upto next corner frequency i.e., 8 rad/s. Step 5: Next basic factor in G ( jw) H ( jw) is

1 (1 + j w / 5)

From M versus log w graph shown in Fig. 8.17 (a), At

w = 8, M = – 18 db

Frequency Response Analysis 197 (0.2, 26)

M(db)

(0.5, 18) (1, 12)

20

–20 db/decade –40 db/decade 0

(7, –6) –20 db/decade

–20

(10, –8) GM = 18 db (8, –18)

(5, –22)

(20, –26)

–60 db/decade –40

–60 wgc = 1.4 rad/s

–80 0.01

0.1

(a)

1

10

w

F(°)

–120

–140 PM = 52°

–160

–180

–200 wpc = 7.3 rad/s –220 0.01

0.1

1

10

w

(b)

Fig. 8.17

(a), (b)

So, at w = 80 (decade system), M = previous M + Net slope (as discussed in section 8.3.1) = – 18 + (– 20 – 40) = – 78 db.

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These points are joined by dotted line on M versus log w graph. This line is made solid further. Step 6: To draw F versus log w graph, firstly phase angle equation is written: 0.05 w w w – tan–1 – tan– 1 F = – 90° + tan– 1 w2 2 0.5 164 The different values of F for different w are: w (rad/s) 1 2 5 10 F (°) –130 –127 –128.4 –190 This F versus log w graph is shown in Fig. 8.17(b). Step 7: From bode plot, gain cross-over frequency, wgc = 1.4 rad/s, phase cross-over frequency, wpc = 7.3 rad/s, GM = 18 db and PM = 52°. As wpc > wgc and GM and PM are both positive, so system is stable. Example 8.3 function as

(Answer)

Draw bode plot and discuss stability for the system having transfer G (s) =

10 s 2 (1 + 0.2 s ) (1 + 0.02 s )

G (s) H (s) =

10 s (1 + 0.2 s ) (1 + 0.02 s )

Solution: Step 1: fi

G ( jw) H ( jw) =

2

10

( j w)

2

(1 + 0.2 j w ) (1 + 0.02 j w )

Here corner frequencies are 5 and 50 rad/s.

=

10 jw 2 ( j w ) ÊÁË 1 + 5 ˆ˜¯

Step 2: The first basic factor associated with G ( jw) H ( jw) is Here,

M = 20 log10 10 – 40 log10 w

At

w = 1, M = 20 db

j wˆ Ê ÁË 1 + 50 ˜¯

10 ( j w )2

At w = 10 (decade system), M = 20 – 40 = – 20 db. These points are plotted on M versus log w graph as dotted line as shown in Fig. 8.18(a). This dotted line is made solid upto first corner frequency i.e., 5 rad/s. 1 Step 3: Next basic factor in G ( jw) H ( jw) is jw ˆ Ê ÁË 1 + 5 ˜¯ From M versus log w graph shown in Fig. 8.18(a), At w = 5, M = – 8 db

Frequency Response Analysis 199

So, at w = 50 (decade system), M = previous M + Net slope (as discussed in section 8.3.1) = – 8 + (– 40 – 20) = – 68 db. These points are joined by dotted line on M versus log w graph. This dotted line is made solid upto second corner frequency i.e., 50 rad/s. M(ab) 40 (1, 20)

20 0

–20 db/decade (5, –8) (10, –20)

–20

–40 db/decade

–40 –60

(50, –68)

–80

–60 db/decade

–100 –120 –140

wgc = 3 rad/s

–160 0.1

1

F(°)

(a)

10

100

(500, – 148) w

10

100

w

–160

–180

PM = –38° –200

–220 wPC = 0 –240 0.1

1 (b)

Fig. 8.18

(a) and (b)

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Step 4: Next basic factor in G ( jw) H ( jw) is

1 jw ˆ Ê ÁË 1 + 50 ˜¯

From M versus log w graph shown in Fig. 8.18 (a), At w = 50, M = – 68 db So, at w = 500 (decade system), M = previous M + Net slope (as discussed in section 8.3.1) = – 68 + (– 60 – 20) = – 148 db. These points are joined by dotted line on M versus log w graph. This line is made solid further. Step 5: To draw F versus log w graph, firstly phase angle equation is written: w w – tan–1 F = – 180° – tan–1 5 50 The different values of F for different w are: w (rad/s) F (°)

0 –180

0.05 –180.5

0.1 –181

1 –192

5 – 230

This F versus log w graph is shown in Fig. 8.18 (b). Step 6: From bode plot, gain cross-over frequency, wgc = 3 rad/s and phase cross-over frequency, wpc = 0 rad/s. GM = • and PM = – 38°. As wpc < wgc, so system is unstable.

8.4

(Answer)

POLAR PLOT

Polar plot of a sinusoidal transfer function G ( jw) H ( jw) is a plot of the magnitude of G ( jw) H ( jw) versus phase angle of G ( jw) H ( jw) on polar coordinates as w is varied from zero to infinity. So, polar plot is the locus of |G ( jw) H ( jw)| and –G ( jw) H ( jw) as w is varied from 0 to •. In other words, it is a plot of magnitude and phase of G ( jw) H ( jw) as • is varied from 0 to •, in a single plot, unlike previous case of bode plot, in which two separate curves were drawn, of magnitude versus frequency and phase angle versus frequency. This is the biggest advantage of drawing polar plot, that it depicts the frequency response characteristics in a single plot. Steps to sketch the polar plot are: Step 1: Determine the transfer function G (s) H (s) of the given system. Replace s by jw to obtain G ( jw) H ( jw). Step 2: Convert G ( jw) H ( jw) into polar coordinates as G ( jw) H ( jw) = |G ( jw) H ( jw)| –G ( jw) H ( jw) = M – F. Step 3: Find M and F at w = 0 and at w = •.

Frequency Response Analysis 201

Step 4: With these values of M and F, sketch polar plot starting at w = 0 and terminating at w = •. Let us consider the simplest transfer function: 1 G (s) H (s) = 1 + sT 1 Step 1: Replace s by jw, G ( jw) H ( jw) = 1 + j wT Step 2:

G ( jw) H ( jw) =

Thus,

M=

1 –– tan– 1 wT 1 + j wT 1

and F = – tan–1 wT

1 + w 2T 2

Step 3: At w = 0; M = 1 and F = 0° (starting point of polar plot) At w = •; M = 0 and F = – 90° (terminating point of polar plot) Step 4: Polar plot is drawn as shown in Fig. 8.19. +90° –270°

1

–180°

w=•

+180°

–90° +270°

Fig. 8.19

8.4.1 (i)

Effect of Adding Poles on Polar Plot

Add a zero-pole (pole at origin) to the above transfer function 1 Thus, new transfer function is G1(s) H1(s) = s (1 + sT ) Replacing s by jw, G 1 ( jw) H1 ( jw) =

1 j w (1 + j wT )

w=0

0° 360°

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Now, M =

1 w 1 + w2 T 2

and F = – 90° – tan–1 wT

At w = 0; M = • and F = – 90° (starting point of polar plot) w = •; M = 0 and F = – 180° (terminating point of polar plot) Polar plot is as shown in Fig. 8.20.

Fig. 8.20

Figure 8.20 shows that the presence of pole at origin shifts the starting and terminating point by – 90°. (ii) Add a non-zero pole (pole having some value other than zero) to the original transfer function. 1 (8.12) G2 (s) H2(s) = (1 + sT1 ) (1 + sT ) 1 Replacing s by jw, G2 ( jw) H2 ( jw) = (1 + j w T1 ) (1 + j wT ) Now, M =

1 2

1 + w T1

2

2

1+w T

2

and F = – tan–1 w T1 – tan–1 wT

At w = 0; M = 1 and F = 0° (starting point of polar plot) w = •; M = 0 and F = – 180° (terminating point of polar plot) Polar plot is as shown in Fig. 8.21. Figure 8.21 shows that presence of non-zero pole to a transfer function shifts its terminating point by – 90°.

Frequency Response Analysis 203

Fig. 8.21

8.4.2

Effect of Adding Zeros on Polar Plot

(i) Add a pure zero or zero-zero (zero at origin) to the original transfer function. s G3(s) H3(s) = (1 + sT ) jw Replacing s by jw, G3 ( jw) H3( jw) = (1 + jwT ) w Now, M = and F = + 90° – tan–1 wT 1 + w2 T 2 At w = 0; M = 0 and F = + 90° (starting point of polar plot) w = •; M =

1

=

1 and F = 0° (terminating point of polar plot) T

1 + T2 w Polar plot is as shown in Fig. 8.22. Figure 8.22 shows that presence of zero at origin to a transfer function shifts its starting and terminating points by + 90°.

(ii) Add a non-zero zero (zero having value other than zero) to the transfer function of equation (8.12). (1 + sT2 ) (8.13) Thus, G4 (s) H4 (s) = (1 + sT1 ) (1 + sT ) (1 + j w T2 ) Replacing s by jw, G4 ( jw) H4 ( jw) = (1 + j w T1 ) (1 + j w T )

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Fig. 8.22

Now, M =

1 + w 2 T22 1 + w 2 T12 1 + w 2 T 2

and F = + tan– 1 wT2 – tan–1 wT1 – tan–1 wT

At w = 0; M = 1 and F = 0° (starting point of polar plot) 1 + T22 w2 = 0 and F = – 90° (terminating point of polar plot) w = •; M = 1 1 2 2 w +T + T1 w2 w2

Polar plot is as shown in Fig. 8.23. Comparing Fig. 8.23 to Fig. 8.21 shows that terminating point shifts by angle of + 90°.

8.4.3 Gain Margin, Phase Margin and Stability from Polar Plot Consider a polar plot of a transfer function, as shown in Fig. 8.24. To find gain margin and phase margin, firstly cross-over frequencies i.e., phase cross-over frequency (wpc) and gain cross-over frequency (wgc) are obtained. Phase cross-over frequency ( pc): The frequency at which polar plot crosses negative real axis is called phase cross-over frequency. It is found out by putting F = – 180° or by putting imaginary part of G ( jw) H ( jw) equal to zero and then solving for w. Gain cross-over frequency ( gc): A circle with radius unity and centre at origin is drawn and cuts the polar plot at say point P, as shown in Fig. 8.24. The frequency corresponding to point P is gain cross-over frequency. It can be found out by putting |G ( jw) H ( jw)| and solving for w.

Frequency Response Analysis 205

Fig. 8.23

Fig. 8.24

Gain Margin:

It is calculated as: GM = 20 log

Phase Margin:

It is calculated as:

1 G ( jw ) H ( jw ) w = w

pc

PM = 180° + – G ( jw) H ( jw)|w = w

gc

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Stability: For stable system, GM and PM > 0 For unstable system, GM and PM < 0 For marginally stable system, GM = PM = 0 The polar plots can be used directly to determine the stability of the system. If the critical point (– 1 + j 0) is within the plot, then the system is unstable. If that point is outside the plot, system is stable. If polar plot touches or passes through that point, system is marginally stable. Example 8.4 For the given transfer function, draw polar plot and thereby find stability. 1 G (s) H (s) = s (1 + 2 s ) (1 + s ) Solution: Step 1: Replace s by jw, 1 G ( jw) H ( jw) = jw (1 + j 2 w ) (1 + j w ) Step 2:

M=

1 w 1 + 4 w2 1 + w2

and F = – 90° – tan–1 2w – tan–1 w

Step 3: At w = 0; M = • and F = – 90° (starting point of polar plot) w = •; M = 0 and F = – 270° (terminating point of polar plot) Step 4: Draw polar plot as in Fig. 8.25.

Fig. 8.25

Step 5: To find wpc; put F = – 180° fi

– 90° – tan–1 2w – tan–1 w = – 180°

Frequency Response Analysis 207

– tan–1 2w – tan–1 w = 90 tan (– tan–1 2w – tan–1 w) = tan 90°

fi fi

tan(tan -1 2 w ) + tan (tan -1 w ) =∞ - tan (tan -1 2 w ) tan (tan -1 w )

fi

2w + w =∞ 1 - 2 w2 3w =∞ 1 - 2 w2

fi fi

1 - 2 w2 =0 3w 1 – 2w2 = 0

fi fi

fi w = 1/√2 = 0.707 rad/s = wpc Step 6: Now, value of a, corresponding to wpc is M|w = w = pc

1 0.707 (1 + 4 / 2)(1 + 1 / 2)

=

2 3/ 2

= 2 / 3 = 0.66

Step 7: So, point (– 1 + j0) is outside the polar plot. Hence, system is stable.

8.4.4 Inverse Polar Plot 1 as w is varied The inverse polar plot of G (jw) H (jw) is polar plot of G ( j w ) H ( j w ) from 0 to •.

Example 8.5

Sketch inverse polar plot of G (s) H (s) =

Solution: Step 1: Step 2:

G ( jw) H ( jw) =

20 s (1 + 2 s ) 20 j w (1 + 2 j w )

(1 + 2 j w ) 1 = 20 j w G ( j w) H ( j w)

Step 3:

M=

1 + 4 w2 20 w

and F = tan–1 2 w – 90°

Step 4: At w = 0; M = • and F = + 90° (starting point of polar plot) w = •; M =

1 +4 2 1 w2 = = and F = 0° (terminating point of polar plot) 20 20 10

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Step 5: Inverse polar plot is drawn in Fig. 8.26. +90° –270°

–180°

w=•

+180°

0.1

0° 360°

w=0 –90° +270°

Fig. 8.26

8.5

NYQUIST STABILITY CRITERION

Nyquist stability criterion is another method to determine the stability of closed loop system by investigating its properties in frequency domain. In previous topics, it was seen that stability of system depends upon the location of roots of characteristics equation in s-plane. For a system to be stable, it was seen that the roots of characteristics equation should lie on the L.H.S. of s-plane. But Nyquist stability criterion is the method that gives stability of the system by checking if there is any pole or zero on R.H.S. of s-plane. This method is based on Principle of Arguments or Cauchy’s Residue Theorem of complex variables. This principle of argument is related to Theory of Mapping. So, before studying Nyquist stability criterion, it is very important to understand Principle of Arguments and Theory of Mapping.

8.5.1

Mapping of Closed Contour

Consider a function y = F (s), s is a real variable. Now, if we have a specified path along which s moves, then we can find the path along which F (s) moves. The path in s-plane is then said to be mapped in F (s) plane. We can take one example to illustrate this. Example 8.6 If F (s) = 2 s + 1, map the s-plane path shown in Fig. 8.27. Solution: Step 1: For point A, s = 2 + 2 j So, F (s) = 2 s + 1 = 2 (2 + 2 j) + 1 = 5 + 4 j

Frequency Response Analysis 209

Fig. 8.27

Similarly, for point B, s = 2 – 2 j; F (s) = 5 – 4 j For point C, s = – 2 – 2 j; F (s) = –3 – 4 j For point D, s = – 2 + 2 j; F (s) = – 3 + 4 j These points are shown in Fig. 8.28 as A¢, B¢, C ¢ and D¢ respectively in F(s)-plane.

Fig. 8.28

So, path in s-plane shown in Fig. 8.27 is mapped in new path in F (s)-plane shown in Fig. 8.28.

8.5.2 Principle of Argument It is a theorem which states that if s-plane contour say C contains Z zeros and P poles of function F (s) within it, then the mapped F (s)-plane contour say P encircles the origin (Z – P) times in the same direction as that of contour C. Also, the total change in the angle of F (s) is equal to – 2 p (Z – P).

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Consider a function F (s) = A

(s - z1 ) (s - z2 ) º (s - p1 ) (s - p2 ) º

where, z1, z2, º are zeros of F (s) p1, p2, º are poles of F (s). Also, assume that only one zero is encircled by contour C in clockwise direction. Vectors a1, a2, a3 and a4 are also shown in Fig. 8.29.

Fig. 8.29

Fig. 8.30

Then, according to the principle of argument theorem, when we move along C, vectors a2, a3 and a4 rotates by zero degree, but vector a1 rotates by an angle – 2 p (as Z = 1 and P = 0). So, the corresponding contour P encircles the origin once in clockwise direction as shown in Fig. 8.30. Now, consider another case, when contour C encircles one pole in clockwise direction as shown in Fig. 8.31. So, when we move along C, vectors b2, b3 and b4 rotate through zero degree, but vector b1 rotate through –2 p. But, as pole in denominator, so net phase rotation is + 2 p (as Z = 0 and P = 1). So, the corresponding contour P encircles the origin once, but in anti-clockwise direction as shown in Fig. 8.32.

Frequency Response Analysis 211 Im

C

b1 b4 b2 Re b3

s-plane

Fig. 8.31

Fig. 8.32

Therefore, in general, if contour C encircles both zeros Z and poles P in clockwise direction, then the corresponding contour P encircles the origin Z times in clockwise direction and P times in anticlockwise direction so that net encirclement of origin is (Z – P) times in clockwise direction and (P – Z) times in anticlockwise direction. The relationship between enclosure of poles and zeros by contour in s-plane and encirclement of origin by contour in F (s)-plane is known as Principle of Argument. Conclusions: If N = P – Z, (i) If N < 0 i.e., Z > P, then P will encircle the origin N times in same direction as that of C. (ii) If N > 0 i.e., Z < P, then P will encircle the origin N times in opposite direction as that of C. (iii) If N = 0 i.e., Z = P, then P will not encircle the origin.

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8.5.3 Nyquist Criterion The general characteristic equation of a closed loop system is given by: D (s) = 1 + G (s) H (s) For stability of this system, the necessary and sufficient condition is that all the roots of the characteristics equation must lie in the L.H.S. of s-plane. The main purpose of Nyquist criterion is to determine whether the system has any poles in R.H.S. of s-plane or not. So, a contour in s-plane is chosen which encloses the entire R.H.S. of s-plane and is called nyquist contour. Suppose there are Z zeros and P poles in R.H.S. of s-plane. If this contour is mapped to PD contour in D (s)-plane, then PD encloses the origin N = (P – Z) times in opposite direction as that of nyquist contour. Now, a feedback system is stable iff there are no zeros of D (s) in R.H.S. of s-plane i.e., iff Z = 0. Thus, for stability N = +P or number of anticlockwise encirclements by PD should be equal to number of poles of D (s) in R.H.S. s-plane. Now,

D (s) = 1 + G (s) H (s)

G (s) H (s) = D (s) – 1 (8.14) Let contour PD in D (s)-plane is mapped to a new contour PGH in G (s) H (s)-plane (known as nyquist plot). PGH can be obtained by shifting PD horizontally to the left by one unit as given by equation (8.14). So, encirclement of origin by PD is equivalent to the encirclement of point (– 1 + j 0) by PGH, as shown in Fig. 8.33. fi

Im PD

PGH

–1 Re

D(s)-plane

Fig. 8.33

Accordingly, statement of Nyquist Criterion becomes “A feedback system is said to be stable if contour PGH of open loop transfer function G (s) H (s) i.e., nyquist plot corresponding to nyquist contour in s-plane encircles the point (– 1 + j 0) in counterclockwise direction the same number of times as the number of poles in R.H.S. of s-plane. For anticlockwise encirclements, N is positive, otherwise N is negative. Stability can be checked using equation N = P – Z and Z must be zero.

Frequency Response Analysis 213

How to map Nyquist contour in s-plane to contour in PGH in G (s) H (s) plane: Case 1:

When no pole at origin K (8.15) G (s) H (s) = 1 + sT Consider Nyquist contour n s-plane covering the entire R.H.S. of s-plane as shown in Fig. 8.34. The contour is having clockwise encirclement and radius R that is approaching infinity. It comprises the following sections: Section AB: Here, s = jw and 0 < w < • Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. Section DA: Here, s = – jw and • > w > 0 (it is mirror image of section AB).

Fig. 8.34

Mapping of each section is done as: Section AB: s = jw and 0 < w < • K 1 + j wT K and f = – tan–1 wT M = |G ( jw) H ( jw)| = 1 + w2 T 2

So, from equation (8.13), G ( jw) H ( jw) = and At

w = 0; M = K and F = 0° w = •; M = 0 and F = – 90° This is same as polar plot shown in Fig. 8.35. So, AB in Nyquist contour is mapped to A¢ B¢ in PGH contour.

Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. K = 0e j q (As R → •) So, G (s) H (s) = 1 + Re j q T

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Fig. 8.35

Thus, BCD is mapped to locus B¢ C ’ D’ having zero radius from – 90° to + 90° through 0° as shown in Fig. 8.36.

Fig. 8.36

Section DA: Here, s = – jw and • > w > 0 It is mirror image of A¢ B¢. So, complete Nyquist plot is shown in Fig. 8.37. As the critical point (–1 + j 0) is not encircled by Nyquist plot. So, N = 0. Also, no pole on R.H.S. of s-plane. So, P = 0. As N = P = 0, so Z = 0. Hence, system is stable. Case 2:

When any pole on jw axis (including at origin) K G (s) H (s) = s (1 + sT )

(8.16)

In this case, Nyquist contour shown in Fig. 8.34 must be modified by, bypassing the jw axis pole with a semicircle of radius l, where l approaches zero, as shown in Fig. 8.38.

Frequency Response Analysis 215

Fig. 8.37

It comprises the following sections: Section AB: Here, s = jw and 0 < w < • Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. Section DE: Here, s = – jw and • > w > 0 (it is mirror image of section AB). Section EFA: s = le j q; l → 0 and q → – 90° to + 90° through 0°.

Fig. 8.38

Mapping of each section is done as: Section AB: s = jw and 0 < w < • So, from equation (9.14), G ( jw) H ( jw) = and

K j w (1 + j w T )

M = |G ( jw) H ( jw)| =

K 2

w 1+w T

2

and F = – 90° – tan–1 wT

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At

w = 0; M = • and F = – 90° w = •; M = 0 and F = – 180°

So, section AB is mapped to A¢B¢ — as shown in Fig. 8.39. Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. K K = = 0 e - j 2 q (as R → •) So, G (s) H (s) = Re j q (1 + Re j qT ) R 2 e 2 j q T Thus, BCD is mapped to locus B¢ C¢ D¢ having zero radius from – 180° to + 180° through 0° as shown in Fig. 8.39. Section DE: DE is mapped to D¢ E¢ which is mirror image of A¢B¢ shown in Fig. 8.39. Section EFA: s = le j q; l → 0 and q → – 90° to + 90° through 0°. K K = = • e – j q (as l → 0) Here, G (s) H (s) = jq jq le (1 + l e T ) l e j q So, EFA is mapped to E¢F¢A¢ with radius • from + 90° to – 90° through 0°. Therefore, complete Nyquist plot is shown in Fig. 8.39.

Fig. 8.39

As the critical point (– 1 + j 0) is not encircled by nyquist plot, so N = 0. There is no pole on R.H.S. of s-plane. So, P = 0. As N = P = 0. So, Z = 0. Thus, system is stable.

8.5.4 Calculation of Gain Margin and Phase Margin and Stability From Nyquist Plot These can be calculated by same method as described for polar plots in section 8.4.3.

Frequency Response Analysis 217

Example 8.7

Consider a unity feedback system with open loop transfer function (4 s + 1) . Comment on stability using stability criterion. G (s) H (s) = 2 s (s + 1) (2 s + 1) Solution: Step 1: As there are two poles at origin, Nyquist contour under consideration is shown in Fig. 8.38. Step 2: Section AB: s = jw and 0 < w < • So,

G ( jw) H ( jw) =

and

4jw + 1 - w ( j w + 1) (2 j w + 1) 2

M = |G ( jw) H ( jw)| =

16 ω 2 + 1 ω

2

2

2

ω +1 4ω +1

and F = tan–1 4 w

– 180° – tan w – tan 2 w At w = 0; M = • and F = – 180° w = •; M = 0 and F = – 270° So, Section AB is mapped to A¢ B¢ — as shown in Fig. 8.40. Step 3: Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. –1

So, G (s) H (s) =

–1

4 Re j q + 1 4 Re j q 2 = = = 0 e - j 3 q (As R → •) R 2 e +2 j q ( Re j q + 1) (2 Re j q + 1) 2 R 4 e 4 j q R 3 e 3 j q

Thus, BCD is mapped to locus B¢ C¢ D¢ having zero radius from – 270° to + 270° through 0° as shown in Fig. 8.43. Step 4: Section DE: DE is mapped to D¢ E¢ which is mirror image of A¢ B¢, shown in Fig. 8.40. Step 5: Section EFA: s = le j q; l → 0 and q → – 90° to + 90° through 0°. 4 l e jq + 1 1 = 2 2 j q = • e – 2 j q (as l → 0) Here, G (s) H (s) = 2 2 j q jq 2 jq l e (1 + l e ) (1 + 2 l e ) l e So, EFA is mapped to E¢ F¢A¢ with radius • from + 180° to – 180° through 0°, shown in Fig. 8.40. Step 6: Complete Nyquist plot is shown in Fig. 8.40. Step 7: To find the position of point Q, firstly find wpc. To find wpc , put imaginary part of G ( jw) H ( jw) = 0. 4jw + 1 10 w 2 + 1 + j (w - 8 w 3 ) = Now, G ( jw) H ( jw) = 2 ( j w ) ( j w + 1) (2 j w + 1) - w 2 (1 + w 2 ) (1 + 4 w 2 ) So, w – 8 w3 = 0 fi wpc = Thus,

1 8

= 0.353 rad/s

Q = M w pc = 0.353 =

16 (0.353)2 + 1 (0.353)2 (0.353)2 + 1 4 (0.353)2 + 1

= 10.7

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Fig. 8.40

Step 8: From position of point Q, it is seen that critical point (– 1 + j 0) is encircled two times in clockwise direction. So, N = – 2. Here, P = 0 as there are no poles in R.H.S. of s-plane. Now, N = P – Z fi Z = P – N = 0 + 2 = + 2 ≠ 0. Thus, system is unstable. (Answer) K . Example 8.8 Draw the Nyquist plot and discuss stability for G (s) H (s) = s (s – 4) Also, find stability using gain margin and phase margin, for K = 20. Solution Step 1: As there is one pole at origin, Nyquist contour under consideration is as shown in Fig. 8.38. Step 2: The Nyquist contour has the following sections and these are mapped to Nyquist plot as shown in Fig. 8.41.

Fig. 8.41

Frequency Response Analysis 219

Section AB: So,

s = jw and 0 < w < • K G ( jw) H ( jw) = j w ( jw - 4) K

and

M = G ( j w) H( j w) =

and

F = – 90° – 180° + tan–1 w/4 As – tan–1

w w 2 + 16 w w = – 180° + tan–1 4 –4

At

w = 0; M = • and F = – 270° w = •; M = 0 and F = – 180° So, section AB is mapped to A¢ B¢ as shown in Fig. 8.41.

Step 3: Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. K K = 2 2 j q = 0 e - j 2 q (as R → ∞) So, G (s) H (s) = jq jq Re ( Re - 4) R e Thus, BCD is mapped to locus B¢C ¢D¢ having zero radius from – 180° to + 180° through 0° as shown in Fig. 8.41. Step 4: Section DE: DE is mapped to D¢ E¢ which is mirror image of A¢ B¢, shown in Fig. 8.41. Step 5: Section EFA:

s = l e j q; l → 0 and θ → – 90° to + 90° through 0°. K K = • e – j (q + 180) (as l → 0 and due to – 4) = Here, G (s) H (s) = l e j q ( l e j q - 4) - 4 l e j q So, EFA is mapped to E¢ F ¢A¢ with radius ∞ from – 90° to – 270° through – 180°, shown in Fig. 8.41. Step 6: As the plot encircles the critical point (– 1 + j 0) once in clockwise direction, so N = – 1. As there is one pole on R.H.S. of s-plane, so P = 1. Now, N = P – Z fi Z = P – N = 1 – (– 1) = 2 ≠ 0. So, the system is unstable. (Answer 1) Step 7: To find wpc, put imaginary part of G ( jw) H ( jw) = 0. Now, So, Step 8:

- K w2 + j 4 Kw K = jw ( j w - 4) w 2 (16 + w 2 )

G ( jw) H ( jw) =

4 Kw = 0 fi wpc = 4 rad/s w (16 + w 2 ) w 16 + w 2 1 = 20 log 10 Gain margin = 20 log10 = 20 M w = w pc 2

1.07 db. Step 9: To find wgc, put M = 1. So,

20 w 16 + w

2

= 1 fi 20 = w 16 + w 2

= 20 log 10 w pc = 4

32 = 5

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On solving,

w2 = 13.5 fi wgc = 3.67 rad/s

o 3.67 = – 47.5°. Step 10: Phase margin = - 180 + f w = w gc = +180° + – 90° – 180° + tan–1 4

As phase margin is negative, system is unstable.

(Answer 2) 2

Example 8.9 Construct Nyquist plot for G (s) H (s) =

0.1 (1 + s ) . Comment on stability. s3

Solution: Step 1: As there are three poles at origin, Nyquist contour chosen is shown in Fig. 8.38. Step 2: Section AB:

s = jw and 0 < w < • 0.1 (1 + j w )2 So, G ( jw) H ( jw) = ( j w )3 0.1 (1 + w 2 ) and F = – 270° + 2 tan–1 w Here, M = |G ( jw) H ( jw)| w3 At w = 0; M = • and F = – 270° w = •; M = 0 and F = – 90° So, section AB is mapped to A¢ B¢ as shown in Fig. 8.42.

Fig. 8.42

Step 3: Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. So,

G (s) H (s) =

0.1 (1 + Re j q )2 R 2 e 2 j q 1 = 3 3 jq = = 0 e - j q (As R → •) (Re j q )3 R e Re j q

Thus, BCD is mapped to locus B¢C¢D¢ having zero radius from – 90° to + 90° through 0° as shown in Fig. 8.42. Step 4: Section DE: DE is mapped to D¢E¢ which is mirror image of A¢ B¢, shown in Fig. 8.42.

Frequency Response Analysis 221

Step 5: Section EFA: s = l e j q; l → 0 and q → – 90° to + 90° through 0°. Here,

G (s) H (s) =

0.1 (1 + l e j q )2 1 1 = = 3 3 j q = • e -3 j q (As l → 0) jq 3 jq 3 (l e ) (l e ) l e

So, EFA is mapped to E¢ F¢ A¢ with radius ∞ from + 270° to – 270° through 0°, shown in Fig. 8.42. Step 6: To find wpc, put F = – 180° + 2 tan–1 w – 270° = – 180°

fi

2 tan–1 w = 90°

fi

tan–1 w = 45° fi wpc = 1 rad/s

fi

Step 7: Point A = M w pc = 1 =

0.1 (1 + 12 ) = 0.2 (1)3

Step 8: So, encirclements of (– 1 + j 0) are zero i.e., N = 0. As there are no poles on R.H.S. of s-plane, P = 0. So, Z = 0. Thus, system is stable. (Answer) K (1 + 2 s ) . Find K for gain margin = 3 db. For this Example 8.10 G (s) H (s) = s (1 + s ) (1 + s + s 2 ) K, find wgc and phase margin. Solution: Step 1: As there is one pole at origin, Nyquist contour chosen is shown in Fig. 8.38. Step 2: Section AB: s = jw and 0 < w < • So,

G ( jw) H ( jw) =

Here,

K (1 + 2 jw ) ( jw ) (1 + jw ) (1 + jw - w 2 )

M = G ( jw ) H ( jw ) =

K 1 + 4 w2 w 1 + w2

(1 - w 2 )2 + w 2

F = + tan–1 2w – 90° + tan–1 w – tan–1 At

w 1 - w2

w = 0; M = • and F = – 90°

1 1 w2 w = 0 and f = - 90∞ - tan -1 2 2 1 -1 (1 - w ) w2 +1 w

K 4+ w = •; M =

w2 1 +

1 w2

= – 90° – tan–1 (0/–1) = – 90° – 180° + tan–1 0 = – 270°.

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So, section AB is mapped to A¢ B¢ as shown in Fig. 8.43.

Fig. 8.43

Step 3: Section BCD: Here, s = Re j q, R → • and q → + 90° to – 90° through 0°. K (1 + 2 Re j q ) 2 Re j q 1 = = = 0 e -3 j q (as R → •) So, G (s) H (s) = ( Re j q ) (1 + Re j q ) (1 + Re j q + R 2 e 2 j q ) R 4 e 4 j q R 3 e 3 j q

Thus, BCD is mapped to locus B¢C¢D¢ having zero radius from – 270° to + 270° through 0° as shown in Fig. 8.43. Step 4: Section DE: DE is mapped to D¢E¢ which is mirror image of A¢B¢, shown in Fig. 8.43. Step 5: Section EFA: s = l e j q; l → 0 and q → – 90° to + 90° through 0°. 1 = • e - j q (as l → 0) Here, G (s) H (s) = l e jq So, EFA is mapped to E¢ F¢A¢ with radius • from + 90° to – 90° through 0°, shown in Fig. 8.43. Step 6: To find To find wpc, put imaginary part of G ( jw) H ( jw) = 0. So, – w (w – 2 w) + (1 + 2 w)2 (1 – w2) = 0 fi wpc = 1.17 rad/s Step 7: Gain Margin = 3 db fi

3 = 20 log10

1 fi K = 0.614 1.152 K

Step 8: To find wgc, put M = 1 fi wgc = 0.97 rad/s Step 9: Phase margin = – 180° + f|w = w

gc

(Answer 1) (Answer 2)

Frequency Response Analysis 223

= + 180° + tan– 1 (2 × 0.97) – 90° – tan–1 (0.97) – tan–1

8.6

0.97 = 22.1° 1 – 0.97 2 (Answer 3).

NON-MINIMUM PHASE SYSTEMS, ALL PASS SYSTEMS, MINIMUM PHASE SYSTEMS

The transfer function with all poles and zeros in the L.H.S. of s-plane, are called minimum phase transfer functions, and such systems are minimum phase systems. Now, consider another transfer function, having a pole-zero pattern which is antisymmetric about imaginary axis i.e., for every pole in L.H.S. of s-plane, there is a zero in its mirror image position. For example: 1 - j wT G1( jw) = 1 + j wT Its pole-zero behavior is shown in Fig. 8.46. This system has a unity magnitude at all frequencies and phase angle (–2 tan–1 wT) varies from 0 to –180° as w is varied from 0 to •. The systems which have this property of unit magnitude at all frequencies with special anti-symmetric pole-zero patterns are called all-pass systems. Consider another systems, which have poles in L.H.S. of s-plane and zeros in both L.H.S. and R.H.S. of s-plane (poles are not permitted to lie in R.H.S. of s-plane, otherwise system would become unstable). Such transfer function is:

Fig. 8.44

G2 ( jw) =

Fig. 8.45

Fig. 8.46

1 - j wT (1 + j w T1 )(1 + j w T2 )

Its pole-zero pattern is shown in Fig. 8.44. Again,

Ê ˆ Ê 1 - j wT ˆ 1 + j wT = G3 ( jw) G1( jw) G2 ( jw) = Á Ë (1 + j w T1 ) (1 + j w T2 ) ˜¯ ÁË 1 + j w T ˜¯

The pole-zero pattern for G3( jw) is shown in Fig. 8.45.

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It is clear that both G2( jw) and G3( jw) have identical magnitude versus frequency relationship but different phase angle versus frequency relationship, shown in Fig. 8.47. G3( jw) has smaller range of phase angle than that of G2 ( jw). So, there is no unique relationship between magnitude and phase of G2 ( jw) and the phase curve can be altered without altering the magnitude curve by addition of all-pass transfer function. Such a function which has one or more zeros in R.H.S. of s-plane is called non-minimum phase transfer function and such system is non-minimum phase system.

Fig. 8.47

Exercise 1. Define the terms: (a) Gain cross-over frequency (b) Phase cross-over frequency (c) Gain margin (d) Phase margin 2. Explain the correlation between frequency domain analysis and time domain analysis of a second order system. Use equations and graphs to show the correlation. 800 (s + 2) . Also 3. Draw bode plot for the following transfer function: G (s) = 2 s ( s + 10) ( s + 40) discuss its stability. 4. Discuss all the frequency domain specifications of a second order system. 5. Sketch polar plot for the following system:

G (s) H (s) =

K (s + 1) s (s + 2) (s + 3) 2

6. The open loop transfer function of a control system is:

G (s) =

K s (s + 2) (s + 10)

Frequency Response Analysis 225

7. Determine the values of K so that the system may be stable with (a) Gain margin equal to 6 db. (b) Phase margin equal to 45°. 8. Discuss principle of argument in relation to Nyquist criterion. 60 . Using Nyquist criterion, 9. Consider transfer function G (s) H (s) = (s + 1) (s + 2) (s + 5) determine whether the closed loop system is stable or not? Use gain margin and phase margin.

Objective Type Questions 1. The octave frequency range is w1 =4 (a) w2 w1 (c) =2 w2

w1 =8 w2 w1 (d) = 10 w2

(b)

2. The initial slope of Bode plot is an indication of (a) Gain margin (b) Phase margin (c) Type of system (d) Order of system 1 3. The function has slope of 1 + 8s (a) + 40 db/decade (b) – 40 db/decade (c) + 20 db/decade (d) – 20 db/decade 4. The Nyquist plot obey’s (a) Principle of argument (b) Principle of motion (c) Both (a) and (b) (d) None of these 5. A system has transfer function as

(a) 5 and 2 (c) 50 and 20

100 (1 + 0.5 s ) . The corner frequencies are: (1 + 0.2 s )

(b) 0.5 and 0.2 (d) 500 and 200

6. The gain cross-over frequency is the frequency at which |G ( jw) H ( jw)| is equal to 1 (a) 0 (b) 2 (c) 1 (d) 2 7. The phase shift of second order system with transfer function 1/s2 is (a) + 180° (b) – 180° (c) 0° (d) 90°

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8. The Nyquist plot of G ( jw) H ( jw) for a closed loop control system passes through (– 1 + j 0) point in GH plane. The gain margin of such a system in db is (a) greater than zero (b) zero (c) infinite (d) less than zero ks + 1 9. Consider a system with G (s) = . The value of k such that the system has s2 gain margin of p/4, is approximately equal to (a) 1.4 (b) 3.4 (c) 0.4 (d) 0.84 3 e– 2 s 10. The open loop transfer function of a unity feedback system is . The gain s (s + 2) and phase cross-over frequencies in rad/s are, respectively

(a) 0.632 and 0.485 (b) 0.632 and 1.26 (c) 0.485 and 0.632 (d) 1.26 and 0.632 11. The gain margin and phase margin for the system given in 9.10 are (a) – 7.09 db and 87.5° (b) 7.09 db and 87.5° (c) 7.09 db and -87.5° (d) -7.09 db and -87.5° 12. The Nyquist plot of G ( jw) H ( jw) for a closed loop control system encloses point (–1 + j 0) in GH plane. The gain margin of such a system in db is (a) less than zero (b) zero (c) infinite (d) greater than zero 13. In the Bode plot of unity feedback control system, the value of phase of G ( jw) H ( jw) at the gain cross-over frequency is – 125°. The phase margin of the system is (a) – 55° (b) 55° (c) – 125° (d) – 125° 14. For the asymptotic Bode plot shown below, the system transfer function can be

(a)

10 s + 1 0.1 s + 1

(b)

100 s + 1 0.1 s + 1

(c)

100 s + 1 10 s + 1

(d)

0.1 s + 1 10 s + 1

Frequency Response Analysis 227

15. For a system to be stable: (a) GM and PM should be positive. (b) GM and PM should be negative. (c) GM and PM should be equal to zero. (d) None of these. 16. For a system to be stable: (a) wpc > wgc (b) wpc < wgc (d) None of these. (c) wpc = wgc 17. The gain margin is the reciprocal of gain at frequency at which phase angle is (a) 0° (b) 180° (c) – 180° (d) 90° 18. The phase cross-over frequency is the frequency at which phase curve crosses (a) 0° (b) 90° (c) – 180° (d) 180° 19. A system has 12 poles and 4 zeros. The slope of its highest frequency asymptote in its M versus log w plot is: (a) – 240 db/decade (b) + 240 db/decade (c) – 200 db/decade (d) + 200 db/decade s is subjected to a sinusoidal input. 20. A system with transfer function G (s) = (1 + s ) In steady-state, the phase angle of output relative to the input at w = 0 and w = • will be respectively (a) 0° and 90° (b) 0° and 0° (c) 90° and 0° (d) 90° and – 90° 21. A minimum phase unity feedback system has a bode plot with a constant slope of – 20 db/decade for all frequencies. What is the value of maximum phase margin for the system? (a) 0° (b) 90° (c) 180° (d) – 90°

Answers 1. 6 11. 16. 21.

(c) (a) (b) (a) (b)

2. 7. 12. 17.

(c) (b) (a) (c)

3. 8. 13. 18.

(d) (b) (b) (c)

4. 9. 14. 19.

(a) (d) (a) (c)

5. 10. 15. 20.

(a) (d) (a) (c)

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MATLAB Programs P1. A unity feedback control system has G (s) = the Bode plot.

40 and H (s) = 1. Draw s (s + 2) (s + 5) 2

Program: n=[40]; d1=conv([1 0 0],[1 2]); d=conv(d1,[1 5]); G=tf(n,d) GH=G*1; bode(GH)

% Numerator of transfer function % % % %

Denominator of transfer function Forward path transfer function Open loop transfer function To draw bode plot

Execution:

Fig. F1

50 . Also find, s (1 + 0.25 s ) (1 + 0.1 s ) gain cross-over frequency, phase crossover frequency, gain margin and phase margin.

P2. Draw Bode plot for the transfer function G (s) =

Frequency Response Analysis 229

Program: n=[50] ; d1=conv([1 0],[0.25 1]); d=conv(d1,[0.1 1]) ; G=tf(n,d); margin(G)

%Numerator of transfer function %Denominator of transfer function %Transfer function %To draw bode plot and to find all the specifications

Execution:

Fig. F2

P3. Draw the Nyquist plot for the system having transfer function 50 . G (s) = (s + 1) (s + 2) (s + 10)

Program: n=[50]; d=conv([1 1],conv([1 2],[1 10]));

%Numerator of transfer function %Denominator of transfer function

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GH=tf(n,d); nyquist(GH)

%Transfer function %To draw nyquist plot

Execution:

Fig. F3

9

Control System Components

CHAPTER

The general block diagram of automatic control system is discussed in detail in section 1.6. The various components in automatic control system are error detectors, control components and feedback elements. The detailed theory of all these components and their examples are given in this chapter. The various components discussed in this chapter are servomotors, potentiometers, synchros, stepper motors and tachogenerators. The basic principle, operation and applications are given for all these components.

9.1

SERVOMOTORS

In section 1.7.1, it was discussed that ‘servo’ is the technical name for word ‘position’. Servosystems are the systems that control the position or time derivative of position i.e., velocity or acceleration. So, servomotors are the motors that convert the electrical signal into movement of shaft. Classification of servomotors is as follows:

9.1.1

AC Servomotors

AC servomotor is a 2-phase servomotor that is similar to 2-phase induction motor, but it has some specific design parameters. The characteristics of AC servomotors are: (i) It has small diameter to length ratio. (ii) It has minimum moment of inertia. (iii) It has maximum acceleration. (iv) It is very rugged, robust and reliable. (v) Its power range is between fractions of watts to hundred of watts. (vi) Its torque speed characteristics must be linear. Figure 9.1 shows the schematic diagram of 2-phase AC servomotor.

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Fig. 9.1

There are two stator windings that are supplied by a 2-phase power supply. In the view of non-availability of 2-phase power supply, reference winding is connected to 1-phase power supply, frequency of which is normally 50 Hz or 400 Hz and control winding is connected to 1-phase power supply through capacitor that will produce a phase shift of 90°. So, the output of the amplifier is given to the control winding through a capacitor and that amplifier is connected to same 1-phase power supply. In addition to 90° phase shift in time, the position of the two windings is also in quadrature that means the reference and control windings are placed 90° apart in space. This is due to the fact that torque produced in the 2-phase motor is most efficient if the following two conditions are met: (i) Voltages in the two phases are in time phase. (ii) Axis of the two phases are in space quadrature. Figure 9.2 shows the torque-speed characteristics of a two-phase servomotor at various values of control voltage Ec. . From Figure 9.2, it is clear that torque is a function of speed q and control voltage Ec. So, the equation becomes, . T = K1 q + K2 Ec (9.1) . Negative sign of K1 shows the inverse relationship between torque and q, positive sign of K2 shows the direct relationship between torque and Ec.

Fig. 9.2

Control System Components 233

The mechanical torque-balance equation from Fig. 9.1 is .. . T = Jq + f q where J = moment of inertia of the system f = friction of the system Equating equations (9.1) and (9.2), . .. . – K1 q + K2 Ec = J q + f q

(9.2)

(9.3)

Now, for Fig. 9.1, Ec is the input variable and q is the output variable, so transfer function is calculated as: Taking Laplace transform of equation (9.3), K2 Ec (s) – (K1 + T) s q (s) – s2 T q (s) = 0 fi

q ( s) K2 = 2 Ec ( s) s J + ( K1 + f ) s

fi

K2 ( f + K1 ) q ( s) K = 2 = 2 ma Ec ( s) s J s Tma + s ( f + K1 ) + s

fi

q ( s) K ma = Ec ( s) s ( sTma + 1)

where

(9.4)

Kma = K2/( f + K1) = ac motor gain constant

Tma = J/( f + K1) = ac motor time constant Equation (9.4) gives the transfer function of a two-phase servomotor. Its block diagram is given in Fig. 9.3.

Fig. 9.3

9.1.2

Armature Controlled DC Servomotors

This is the DC servomotor having separately excited field excited by permanent magnets. The control signal from amplifier is applied to armature winding, hence the name armature controlled DC servomotor is given. The schematic diagram is shown in Fig. 9.4. Nomenclature: if = field current (constant) ia = armature winding current (variable) ea = armature voltage supplied by amplifier Ra = resistance of armature winding

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Fig. 9.4

La = inductance of armature winding eb = back emf induced q = angular displacement of shaft T = torque delivered by motor J = moment of inertia of system f = frictional force of system The total torque T delivered by the motor depends upon the product of armature current ia and air gap flux f. Air gap flux is further dependent upon field current, which is a constant. So, T depends upon only armature current ia , i.e., T = Kia

(9.5)

The KVL equation in armature circuit is dia ea = Ra ia + La + eb (9.6) dt When armature rotates, voltage eb proportional to the product of flux and angular velocity is induced in armature. As flux is constant, so dq (9.7) e b = Kb dt Thus equation (9.7) becomes dq dia + Kb ea = Ra ia + La dt dt Taking Laplace transform on both sides, (9.8) Ea(s) = RaIa (s) + sLaIa (s) + sKb q(s) The mechanical torque-balance equation for the system shown in Fig. 9.4 is d2 q dq = Kia (from equation (9.5)) T= J 2 + f dt dt Taking Laplace transform on both sides, (9.9) T (s) = s 2 Jq (s) + sf q (s) = KIa (s) In this system, q (s) is the output variable and Ea (s) is the input variable, so that using equations (9.8) and (9.9), transfer function becomes, q ( s) K = Ea ( s) s [s 2 JLa + s ( fLa + JRa ) + ( fRa + KK b )]

Control System Components 235

Assuming La to be very very small, q ( s) K = 2 Ea ( s) s JRa + s ( fRa + KK b )]

fi

K q ( s) ( f Ra + KK b ) = Ea ( s) s2 Ê JRa ˆ f Ra + KK b ¯ + s Ë

fi

K q ( s) ( fRa + KK b ) = Ea ( s) s2 Ê JRa ˆ fRa + KK b ¯ + s Ë

fi

q ( s) K mda = Ea ( s) s (sTmda + 1)

(9.10)

The final transfer function is given by equation (9.10), where Kmda = DC armature controlled motor gain constant Tmda = DC armature controlled motor time constant

9.1.3 Field Controlled DC Servomotor This is the DC servomotor that has separately excited field supplied through control amplifier. The armature current ia is maintained constant, by supplying the armature through a constant voltage source or by inserting very high resistance in series with the armature. The schematic diagram is shown in Fig. 9.5.

Fig. 9.5

Nomenclature: ef = if = ea = ia = Rf = Lf = Ra =

field voltage field current (variable) armature voltage (constant) armature current (constant) resistance of field winding inductance of field winding resistance of armature winding

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La = inductance of armature winding q = angular displacement of shaft T = torque delivered by motor J = moment of inertia of system f = frictional force of system The torque developed by the motor is the proportional to product of armature current ia and air gap flux f. Air gap flux is further dependent upon field current if , torque equation becomes, T = Kia if Now, ia is also constant. So, T = Kif Taking Laplace transform on both sides, T (s) = KIf (s)

(9.11)

Applying KVL in field winding circuit, di f ef = Rf if + Lf dt Taking Laplace transform on both sides, Ef (s) = sLf If (s) + Rf If (s)

(9.12)

The mechanical torque-balance equation for the system shown in Fig. 9.5 is T= J

d2 q dq +f = Ki f 2 dt dt

Taking Laplace transform on both sides, T (s) = s2 Jq (s) + sf q(s)

(9.13)

Putting equation (9.11) in (9.13), s2 Jq (s) + sf q (s) = KIf (s) (9.14) In this system, q (s) is the output variable and Ef (s) is the input variable, so that using equations (9.12) and (9.14), transfer function becomes, q ( s) K = E f ( s) s ( sL f + R f ) (sJ + f )

fi

K f Rf q ( s) = E f ( s) Ê sL ˆ s Á f R + 1˜ Ê s J f + 1ˆ ¯ Ë ¯Ë f

fi

K mdf q ( s) = E f ( s) s sTmdf + 1 (sTmds + 1)

(

)

(9.15)

Control System Components 237

So, the final transfer function of the DC field controlled servomotor is given by equation (9.15), where Kmdf = DC field controlled motor gain constant, Tmdf = DC field controlled field time constant, Tmds = DC field controlled shaft time constant Limitations of field controlled motor are: (i) Due to the presence of back emf in the armature circuit, ia is more difficult to maintain constant as compared to if . (ii) Its efficiency is low. (iii) Lf is not negligible, so that transfer function is third order and hence, its stability is poorer. But overall, DC servomotors are more efficient than ac servomotors. Also, DC servomotors are used where higher shaft power output is required. In addition to this, linearity of characteristics is achieved more easily in DC servomotors. Hence, it is concluded that armature controlled DC servomotors are the most efficient and useful servomotors.

9.2

ERROR DETECTORS

As discussed in section 1.6, these are the components of a control system, which compares the output signal to the reference signal and produces the error signal, which in amplified form is responsible for the control action. In this chapter, two error detectors are discussed, namely, potentiometers and synchros.

9.2.1

Potentiometers

A potentiometer is a device that converts the linear or angular displacement into voltage signal. When two potentiometers are connected in parallel, with a fixed voltage source, then the complete unit can act as an error detector. This is shown in Fig. 9.6.

Fig. 9.6

The sliding contacts of both the potentiometers i.e., 1 and 2 are brought out at a and b. The voltage across a and b i.e., VE is the voltage corresponding to the difference in the positions of sliding contacts at 1 and 2 respectively. If the position of sliding contacts

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is same at 1 and 2 i.e., if x1 = x2, then VE = 0. If x1 ≠ x2, VE has some value. If x1 > x2, then VE will be positive, if x1 < x2, VE will be negative. So, VE is directly proportional to the difference in the levels of sliding contacts at 1 and 2 i.e., at x1 and x2. The transfer characteristics of a potentiometer as an error detector are shown in Fig. 9.7.

Fig. 9.7

In practical system, in place of linear potentiometers, circular wire potentiometers are used as shown in Fig. 9.8. In these potentiometers, linear x1 and x2 are replaced by angular displacement q1 and q2 respectively.

Fig. 9.8

Accordingly, the block diagram representation of this system is shown in Fig. 9.9.

Fig. 9.9

The transfer function becomes, VE ( s) = Ke q e ( s) where Ke = constant of potentiometer error detector qe = change in the position of sliding contact

Control System Components 239

Such an error detector has an advantage that errors due to temperature effects are avoided because both the wires are affected by temperature to the same extent.

9.2.2

Synchros

Synchro is a transducer that generates voltage signal proportional to the angular position of shaft. The first kind of synchro is called synchro transmitter. Synchro transmitter: Construction-wise, it is similar to three-phase alternator. It consists of stator, made up of laminated Si steel, carrying balanced three phase windings S1, S2 and S3. The three-phase windings are star connected as shown in Fig. 9.10. In the stator core, rotor is placed having single winding, whose terminals are brought out and are connected to the slip rings for proper and unlimited rotations. Synchro can be considered same in action as that of transformer, in which rotor acts as primary winding and stator acts as secondary winding.

Fig. 9.10

A single-phase ac supply is given to the rotor winding, which is given by equation (9.16). (9.16) Er = Em sin wt The applied rotor voltage causes the magnetizing current to flow in the rotor winding. So, sinusoidal varying magnetic flux is produced and distributed in the air gap. Now, due to transformer action, voltages are induced in each stator winding which is proportional to the cosine of angle between stator winding and rotor winding axes. Let at any instant, the rotor coil make an angle q with stator coil S1. So, each phaseto-neutral voltage is given by ES 1 n = KEm sin wt cos q

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ES 2 n = KEm sin wt cos (w – 120°) ES 3 n = KEm sin wt cos (q – 240°) where K is a constant depending upon the turns ratio and magnetic coupling between the two windings. The terminal voltages are ES 1 S 2 = ES 1 – ES 2 = KEm sin wt [cos q – cos (q – 120°)] fi Similarly,

ES 1 S 2 = KEm sin wt sin (q + 120°)

(9.17)

ES 2 S 3 = KEm sin wt sin q

(9.18)

and ES 3 S 1 = KEm sin wt sin (q + 240°) (9.19) When q = 0, i.e., when rotor is in line with stator winding S1, then ES 1 n is maximum and ES 2 S 3 is zero. This position of rotor (in line with S1) is known as electrical zero and is taken as reference for angular position of rotor. According to equations (9.17), (9.18) and (9.19), the input to the synchro transmitter is angular position of rotor q and output is three single phase voltages of stator. Synchro transmitter-control transformer pair: Control transformer is another kind of synchro. The rotor of control transformer is of cylindrical shape. When synchro transmitter is used in combination with control transformer, then this pair acts as error detector, as shown in Fig. 9.11.

Fig. 9.11

The stator windings of transmitter are electrically connected to stator windings of control transformer. So, current having same magnitude but opposite direction will flow in the stator winding of control transformer. The induced voltage in the rotor of control transformer i.e., e (t) depends upon the angle of rotor of control transformer and angle of magnetic field which further depends upon the angular position of transmitter rotor.

Control System Components 241

Here, e (t) = KVm sin wt cos a (9.20) where a is net displacement between two rotors. When a = 90°, e (t) = 0. This position of control transformer rotor is known as electrical zero of control transformer, and is taken as reference position of control transformer. Let at any instant, the rotor of transmitter rotate through angle q from its electrical zero position and rotor of control transformer rotate through an angle f from its electrical zero position. So, a = 90° – q + f (9.21) Putting equation (9.21) into equation (9.20), e (t) = K ¢ Vm sin wt cos (90° – q + f) fi

e (t) = K ¢ Vm sin wt sin (q – f)

(9.22)

Equation (9.22) shows that the induced voltage in rotor of control transformer depends upon the difference in the rotor position of the synchro transmitter and synchro control transformer. So, e (t) directly gives the difference in the angular position of the two rotors and hence the combination of synchro transmitter and synchro control transformer acts as an error detector.

9.3

TACHOGENERATORS

It is also known as tachometer. It is an electromechanical transducer that converts mechanical energy into electrical energy. Mechanical energy is in the form of voltage signal and electrical energy is in the form of voltage signal. There are two types of tachogenerators: DC tachogenerators and AC tachogenerators.

9.3.1 DC Tachogenerators It is a small low voltage DC tachogenerator having linear characteristics. It consists of stator and rotor assembly. Magnetic field is provided by permanent magnets. Armature is provided with commutator and brush assembly. The output voltage of tachogenerator is proportional to the shaft speed. Magnitude of voltage is decided by the magnitude of speed of shaft and polarity of voltage is dependent on the direction of rotation of shaft. The schematic diagram is shown in Fig. 9.12.

Fig. 9.12

The output voltage v (t) is given by v (t) μ d q/dt

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where q is angular displacement of load shaft in radians. fi

v (t) = K1 dq/dt

(9.23)

Taking Laplace transform on both sides, V (s) = sK1 q (s) fi

Transfer function =

V ( s) = sK 1 q ( s)

(9.24)

Equation (9.24) gives the transfer function of DC tachogenerator.

9.3.2 AC Tachogenerators It is similar to two-phase induction motor, having two-phase stator windings at right angles to each other. Rotor is squirrel cage type. It is shown in Fig. 9.13.

Fig. 9.13

AC supply is applied to the reference winding, having fixed amplitude and frequency. The voltage across output winding is the final output voltage. When AC is applied to reference winding, reference flux f1 is produced in it. Rotor is in the form of drag type cup. When it rotates, eddy currents are produced, generating rotor flux f2. f2 is then added to f1. Due to this change in the total flux, voltage is induced in the output winding of stator that is proportional to the rotor speed. The input-output relationship for AC tachogenerator is given by (9.25) v (t) μ d q/dt fi v (t) = K2 dq/dt Taking Laplace transform on both sides, V (s) = sK2 q (s) fi

Transfer function =

V ( s) = sK 2 q ( s)

(9.26)

Equation (9.26) gives the transfer function of AC tachogenerator. Its block diagram is shown in Fig. 9.14.

Fig. 9.14

Control System Components 243

Tachogenerators can be used as feedback transducers in various control systems like position control system and speed control system. This is diagrammatically shown in Fig. 9.15.

Fig. 9.15

9.4

STEPPER MOTORS

These are the motors that convert electrical energy in the form of train of pulses into mechanical energy in the form of incremental movements of rotor shaft. These incremental movements may be the shaft linear or angular movements. For one pulse, the rotor shaft moves through one angle, which is known as step angle, expressed in degrees. There are two basic types of stepper motors: permanent magnet stepper motor and variable reluctance stepper motor.

9.4.1 Permanent Magnet Stepper Motor It is a motor consisting of stator and rotor. Stator is multiplier having excitation winding around the poles. The rotor is made up of some ferromagnetic material, which is permanently magnetized and hence this motor is called permanent magnet stepper motor. Figure 9.16 shows such type of motor having 4-pole stator. Figure 9.17 shows the driving circuit of such a motor.

Fig. 9.16

Fig. 9.17

When phase AA1 is excited by closing switch SW1, north pole is created in phase A. Rotor starts to rotate in such a way so as to adjust its magnetic axis with the magnetic axis of stator, as shown in Fig. 9.18(a).

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Fig. 9.18

Now, phase B is excited with switching on SW2 and switching off of SW1. So, north pole is created in phase B and rotor rotates through angle 90°, so as to adjust its magnetic axis with the magnetic axis of stator as shown in Fig. 9.18(b). Similarly, if phases C and D are excited, the rotor rotates through angle 90° in clockwise direction by excitation of each phase. Hence, step angle (a) is given by equation (9.27). 360∞ (9.27) mP where m is number of stacks in stator and P is number of poles. Here, m = 1 and P = 4. So, a = 90° This step angle can be reduced by increasing the number of poles on stator. Major disadvantage of this type of stepper motor is that their size becomes very large. Also, there is restriction on the achievement of smallest step angle. These limitations are overcome by using variable reluctance stepper motor.

a=

9.4.2

Variable Reluctance Stepper Motor

It also consists of stator and rotor. Stator has a common frame with six salient poles and three phases. Stator is in laminated and in single stack form. Rotor has four salient

Control System Components 245

poles as shown in Fig. 9.19. The relationship between rotor poles and stator poles is given by equation (9.28). Ps (9.28) Pr = Ps ± m where m = number of phases on stator = 3 Ps = number of pole pairs on stator = 3 Thus, Pr can be 4 or 2.

Fig. 9.19

Figure 9.20 shows the basic driving circuit of variable reluctance stepper motor.

Fig. 9.20

When phase AA¢ is excited with switch SW1 closed, it produces a magnetic field. The rotor moves in the direction of minimum reluctance, as shown in Fig. 9.21(a).

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Now, phase BB ¢ is excited with SW2 closed and SW1 opened. Here, rotor will try to attain the position of minimum reluctance and hence rotates through angle 30° in anticlockwise direction as shown in Fig. 9.21(b). Next, phase CC ¢ is excited with SW3 closed and SW2 opened. Again, rotor will try to attain the position of minimum reluctance and hence rotates through angle 30° further in anticlockwise direction as shown in Fig. 9.21(c). A

A

q

q

X C¢

X

B¢

B

C

C¢

X

X

B¢ A¢

A¢ (a)

(b)

B

C

A

C¢

B

X B¢

C

X A¢ (c)

Fig. 9.21

So, if the three phases are excited in a specified sequence successively, then rotor will take 12 steps of 30° each to complete one revolution. For this motor, step angle a is given by equation (9.29). a= For

360∞ mPr

(9.29)

m = 3 and Pr = 4, a = 30°.

This step angle can further be reduced if two phases are excited simultaneously. For example, if AA¢ and BB¢ are excited simultaneously, then step angle becomes 15°. It is also seen that when the phase sequence for exciting the phases is ABCABC…, then the rotor rotates in anticlockwise direction through step angle a. But, if phase sequence is ACBACB…, then the rotor rotates in clockwise direction through same step angle a. The variable reluctance stepper motor has many advantages like simple and low cost, small size, high torque to inertia ratio.

Exercise 1. Explain the working and derive the transfer function of (a) armature controlled DC servomotor (b) field controlled DC servomotor 2. Explain the working and derive the transfer function of AC servomotor.

Control System Components 247

3. Discuss with the help of neat diagrams, the working and applications of potentiometers. 4. Discuss in detail the construction and working principle of synchros. 5. Explain, how synchros can be used as an error detector in conjunction with control transformer. Draw diagrams to support your answer. 6. Elaborate, why rotor of synchro control transformer is made cylindrical in shape? 7. Discuss the construction and working of (a) DC tachometers (b) AC tachometers 8. Write the transfer function equations and draw the block diagrams of (a) DC tachometers (b) AC tachometers 9. Explain various types of stepper motors. 10. Discuss the working of permanent magnet stepper motors with neat and clean diagrams.

Objective Type Questions 1. Potentiometer in a control system can act as (a) error detector (b) actuator (c) amplifier (d) feedback element 2. Stepper motor in a control system can act as (a) error detector (b) actuator (c) amplifier (d) feedback element 3. Synchros in a control system can act as (a) error detector (b) actuator (c) amplifier (d) feedback element 4. Tachogenerator in a control system can act as (a) error detector (b) actuator (c) amplifier (d) feedback element 5. Amplidyne in a control system can act as (a) error detector (b) actuator (c) amplifier (d) feedback element 6. A synchro transmitter consists of (a) salient pole rotor winding excited by a DC source (b) cylindrical rotor winding and a stepped stator excited by pulses (c) three-phase balanced stator winding excited by AC signal and rotor excited by DC signal (d) salient pole rotor winding excited by AC supply and three phase balanced stator winding

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7. The reference winding phase and control winding phase in an AC servomotor are (a) in same phase (b) 180° out of phase (c) 90° out of phase (d) 120° out of phase 8. AC tachogenerator is similar to (a) two-phase induction motor (b) single-phase induction motor (c) three-phase induction motor (d) none of these 9. What will be the step angle in permanent magnet stepper motor if number of stacks on stator is 2 and number of poles is 4? (a) 90° (b) 60° (c) 45° (d) 120° 10. How many poles are required on rotor of variable reluctance stepper motor having 4 poles on stator and 4 phases? (a) 3 or 6 (b) 2 or 4 (c) 4 or 8 (d) 5 or 10 11. Electrical zero position is obtained in synchro transmitter–receiver pair when the rotors of the two are at angle. (a) 0° (b) 60° (c) 90° (d) 45°

Answers 1. (a) 6. (d) 11. (c)

2. (b) 7. (c)

3. (a) 8. (a)

4. (d) 9. (c)

5. (c) 10. (b)

10

Robust Control Systems

CHAPTER

10.1

INTRODUCTION

In the preceding chapters, the design of control systems was based on many assumptions. So, the designed control system will always be an inaccurate representation of real physical system. Some of the reasons for these inaccuracies are: change in the system parameters, neglecting the time delays in various components, changes in the operating point or in the equilibrium position, presence of various noises and neglecting various disturbances. In this chapter, the design and analysis of robust control system is discussed, that assures the system performance in spite of all the above said inaccuracies. These inaccuracies are often called uncertainty in a control system. So, a control system is said to be robust when it exhibits the desired performance even in the presence of various plant uncertainties. Such a robust control system that includes the various noises and external disturbance signals is shown in Fig. 10.1.

Fig. 10.1

10.2

SYSTEM SENSITIVITY

The design problem of robust control system is often referred to as sensitivity design problem. Robustness of a control system is also defined as the sensitivity of the control system to the effects of various neglected parameters like disturbances, noises, time delays, etc.

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A system is said to be robust if (i) It has lower sensitivity (ii) It is stable over the range of parameter variations. In this section, measure of robustness is done in terms of two sensitivities that is system sensitivity and root sensitivity. System sensitivity is defined as the ratio of change of system transfer function to the change in some parameter of the system, for small incremental change. It is given by equation (10.1).

SGT =

∂T ∂G

T

(10.1)

G where T = system transfer function, G = system parameter It is also known as logarithmic sensitivity. With the increased importance of pole-zero approach, it becomes more useful to express the sensitivity in terms of roots of the characteristic equation. Root sensitivity is defined as the ratio of change in the location of roots in s-plane to the change in system parameter. It is given by equation (10.2).

∂r ∂G G where r = roots of characteristic equation, G = system parameter. SKr =

(10.2)

m

Again,

T (s) =

K p (s + z j ) j=1 n

p (s + ri )

i=1

n 1 Now, system sensitivity SGT = ∂ ln K - Â ∂ri ∂ G ln ∂ ln G i = 1 (s + ri )

(10.3)

∂zi =0 Here, it is assumed that zeros zi of T (s) are independent of G, so that ∂ ln G From equations (10.2) and (10.3), n

SGT = - Â SGri i=1

1 (s + ri )

(10.4)

Thus, according to the equation (10.4), the two sensitivities, that is, system sensitivity S TG and root sensitivity SGri are directly related. A system is said to be robust if it exhibits essentially the same response for a wide range of system parameter.

Robust Control Systems 251

10.3

ANALYSIS OF ROBUSTNESS

For the system shown in Fig. 10.1, the objectives of the robust control system are: (i) For a given input R (s), error signal E (s) must be minimum. (ii) For a disturbance Td (s), the output signal C (s) must be small. (iii) Sensor noises N (s) must be small as compared to input signal R (s). Now, sensitivity of the system is given by 1 (10.5) S= 1 + Gc (s) Gs (s) H (s) Gc (s) Gs (s) (10.6) and transfer function T = 1 + Gc (s) Gs (s) H (s) From equations (10.5) and (10.6), S+T=1 Thus, for a robust control system, S must be made small. If disturbance is of additive type, then forward path transfer function becomes Ga (s) = Gs (s) + A (s) where A (s) is the disturbance that is bounded in magnitude. Also, it is assumed that both Ga (s) and Gs (s) have same number of poles in the R.H.S. of s-plane. So, condition for stability becomes, |A ( jw)| < |1 + Ga ( jw)|

for all w

(10.7)

Equation (10.7) gives the condition for stability of robust control system under the condition of additive disturbance. If disturbance is of multiplicative type, then forward path transfer function becomes, Gm (s) = Gs (s) [1+ M (s)] where M (s) is the disturbance that is bounded in magnitude. Again, it is assumed that both Gm (s) and Gs (s) have same number of poles in the R.H.S. of s-plane. So, condition for stability becomes, |A ( jw) < 1 +

1 Gm ( jw )

for all w

(10.8)

Equation (10.8) gives the condition for stability of robust control system under the condition of multiplicative disturbance.

10.4

PID CONTROLLERS

The error detector in a closed loop control system detects any error if present between output and reference input. This error signal is used as a control signal for taking control action through a controller. The block diagram representation is shown in Fig. 10.2.

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Fig. 10.2

Here, error signal = reference signal – feedback signal (or output signal) The relation between error signal and control signal depends upon the type of controller.

10.4.1 Types of Controllers The various types of controllers can be (i) Proportional controller (ii) Integral controller (iii) Derivative controller (iv) Proportional-integral controller (v) Proportional-derivative controller (vi) Proportional-integral-derivative controller The selection of a particular type of controller depends upon the operating conditions and nature of operating system like its weight and size, cost, precision required, reliability required, etc.

10.4.2 Proportional Controller When the relation between the error signal e and output of the controller a is linear, then the controller is proportional controller. Its block diagram is shown in Fig. 10.3(a). Its input-output equation is given by a (t) = Kp e (t) Taking Laplace transform on both sides, A (s) = Kp E (s) Now, transfer function is given by equation (10.10). A (s) = Kp E (s)

(10.9)

(10.10)

10.4.3 Integral Controller When the output of the controller a is dependent upon the integration of error signal e, then the controller is integral controller.

Robust Control Systems 253

Its block diagram is shown in Fig. 10.3(b). Its input-output equation is given by (10.11) a (t) = Ki Ú e (t) dt Taking Laplace transform on both sides, Ki E (s) A (s) = s Now, transfer function is given by equation (10.12). A (s) K i = E (s) s

(10.12)

10.4.4 Derivative Controller When the output of the controller a is dependent upon the first order derivative of error signal e, then the controller is derivative controller. Its block diagram is shown in Fig. 10.3(c). Its input-output equation is given by de (t ) (10.13) a (t) = Kd dt Taking Laplace transform on both sides, A (s) = sKd E (s) Now, transfer function is given by equation (10.14). A (s) = sK d (10.14) E (s)

Fig. 10.3

The effect of increasing the controller constants independently is given in Table 10.1. Table 10.1 Parameter

Rise time

Overshoot

Kp

Decrease

Increase

Ki

Decrease

Increase

Kd

Minor change Decrease

Settling time Small change

Steady-state error

Stability

Decrease

Degrade

Increase

Eliminate

Degrade

Decrease

No effect in theory Improve if Kd small

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10.4.5 Proportional-integral Controller This PI controller is the combination of proportional controller and integral controller. Its block diagram is shown in Fig. 10.4(a). Its input-output equation is given by a (t) = Kp e (t) +

Kp Ti

Ú e (t ) dt

(10.15)

where Ti is the time constant of integral controller. Taking Laplace transform on both sides, Ê 1 ˆ E (s) A (s) = K p Á 1 + sTi ˜¯ Ë

Now, transfer function is given by equation (10.16). A (s) = Kp E (s)

Ê 1 ˆ ÁË 1 + sT ˜¯ i

(10.16)

The characteristics of a PI controller are: (i) Offset of the system i.e., steady state error reduces. (ii) Time constant of the system decreases, hence response becomes faster. (iii) Oscillations in the system response increase, hence maximum overshoot increases. (iv) Order of the system increases, so response becomes more sluggish. (v) Bandwidth of the system decreases.

10.4.6 Proportional-derivative Controller This PD controller is the combination of proportional controller and derivative controller. Its block diagram is shown in Fig. 10.4(b). Its input-output equation is given by de (t ) dt where Td is the time constant of integral controller. Taking Laplace transform on both sides,

a (t) = Kp e (t) + K p Td

(10.17)

A (s) = Kp (1 + sTd) E (s) Now, transfer function is given by equation (10.18). A (s) (10.18) = K p (1 + sTd ) E (s) The characteristics of a PD controller are: (i) Damping of the system increases, hence maximum overshoot decreases. (ii) System becomes more robust. (iii) System response becomes slower.

Robust Control Systems 255

Fig. 10.4

10.4.7 Proportional-integral-derivative Controller This PID controller is the combination of proportional controller, integral controller and derivative controller. Its block diagram is shown in Fig. 10.4(c). Its input-output equation is given by Kp de ( t ) e ( t ) dt + K p Td (10.19) a (t) = Kp e (t) + Ú Ti dt Taking Laplace transform on both sides, Ê ˆ 1 + sTd ˜ E ( s) A (s) = K p Á 1 + sTi Ë ¯

Now, transfer function is given by equation (10.20). Ê ˆ A ( s) 1 = K p Á1 + + sTd ˜ E ( s) sTi Ë ¯

(10.20)

The characteristics of a PID controller are: (i) Offset of the system i.e., steady state error reduces. (ii) Damping of the system increases, hence maximum overshoot decreases. (iii) System becomes more robust. (iv) System response becomes faster. PID controllers are the most widely used controllers as they use the best features of both PI and PD controllers. PID controllers involve three separately constant parameters Kp, Ti and Td. Hence, these controllers are also known as three-term PID controllers.

10.5

TUNING OF PID CONTROLLERS

Tuning of PID controllers is the adjustment of its control parameters i.e., Kp, Ti and Td to the optimum values for the desired control response. The basic requirement for the tuning of these parameters is the stability of the system.

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Tuning of PID controllers is a difficult problem, even though there are only three parameters and in principle is simple to describe, because it must satisfy complex criteria within the limitations of PID control. PID controllers often provide acceptable control using default tunings, but performance can generally be improved by careful tuning. Its performance may be unacceptable with poor tuning. There are various methods for PID tuning. One such basic and important method is Ziegler-Nichols method.

10.5.1 Ziegler-Nichols Method The Ziegler–Nichols tuning method is a heuristic method of tuning a PID controller. It was developed by John G. Ziegler and Nathaniel B. Nichols in the 1940s. This method helps in setting the values of Kp, Ti and Td that will give the stable operation of the system. But sometimes, the method becomes unacceptable as maximum overshoot increases with step input. So, a series of fine tuning is needed in that case until an appropriate and acceptable values of control parameters are obtained. So, Ziegler– Nichols rules give an educated guess of control parameter values and provide the starting point for fine tuning of these parameters. There are further two methods that can be used under Ziegler–Nichols method. (i) First method: This method is applicable when there are neither integrators nor complex conjugate poles in the system. Unit step input is applied to the system and its output response is obtained as shown in Fig. 10.5. It will be a S-shaped graph. This S-shaped graph is characterized by two constants Tde and Tc i.e., delay time and time constant respectively these are also shown in Fig. 10.5. The approximated transfer function for a first order system can be written as C (s ) Ke– Tde S = U (s ) sTc + 1

Fig. 10.5

Robust Control Systems 257

For this system, the Ziegler–Nichols chart is given in Table 10.2. Table 10.2 Ziegler–Nichols method

Kp

Ti

Td

1.2 Tc/Tde

2 Tde

0.5 Tde

(ii) Second method: In this method, firstly Ti is set at • and Td at 0. Then, Kp is increased until the system reaches its limit of stability i.e., it becomes marginally stable. Let this value of critical gain be Kc. The corresponding time period of oscillations is Tc as shown in Fig. 10.6.

Fig. 10.6

According to the values of Kc and Tc , the Ziegler–Nichols chart is given in Table 10.3. Table 10.3 Ziegler–Nichols method

Kp

Ti

Td

0.6 Kc

0.5 Tc

0.125 Tc

These control parameter values can be applied to the ideal and parallel form of PID controller. Ziegler–Nichols method of tuning PID controller can be applied to the control systems having both mathematical model representation and transfer function representation.

Exercise 1. Draw and explain the general block diagram of a robust control system. 2. What is meant by sensitivity of a control system? What is its importance in making a system robust? 3. Discuss system sensitivity and root sensitivity of a robust control system.

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4. Discuss the effect of additive type of disturbance and multiplicative type of disturbance in a robust control system. 5. Explain how control signal is generated in a control system? Discuss different types of controllers. 6. What is meant by tuning of PID controllers? What is its importance? 7. Discuss Ziegler-Nichols method of PID tuning in detail.

Objective Type Questions 1. The input-output equation for proportional controller is

(a) a (t) = Kp e (t) (c) a (t) = K d 2.

de (t ) dt

(b) a (t) = Ki Ú e (t) dt (d) none of these

The input-output equation for integral controller is (a) a (t) = Kp e (t) (b) a (t) = Ki Ú e (t) dt

de (t ) (d) none of these dt 3. The input-output equation for derivative controller is

(c) a (t) = K d

(a) a (t) = Kp e (t)

de (t ) (d) none of these dt According to first method of Zeiglar-Nichols rules, Ti is (a) 2 Tde (b) 1.2 Tde (c) 0.2 Tde (d) 2.2 Tde According to first method of Zeiglar-Nichols rules, Td is (a) 2 Tde (b) 0.5 Tde (c) 0.2 Tde (d) 5 Tde According to second method of Zeiglar-Nichols rules, Td is (a) 2 Tc (b) 0.5 Tc (c) 0.125 Tc (d) 1.5 Tc According to first method of Zeiglar-Nichols rules, Ti is (a) 2 Tc (b) 1.5 Tc (c) 1.2 Tc (d) 0.5 Tc According to first method of Zeiglar-Nichols rules, Kp is (a) 1.2 Kc (b) 1.5 Kc (c) 0.6 Kc (d) 0.5 Kc

(c) a (t) = K d

4.

5.

6.

7.

8.

(b) a (t) = Ki Ú e (t) dt

Robust Control Systems 259

Answers 1. (a) 6. (c)

2. (b) 7. (d)

3. (c) 8. (c)

4. (a)

5. (b)

11

Sampled Data Control System

CHAPTER

11.1

INTRODUCTION

In the control systems dealt so far, the signal at every point in the system is a continuous function of time, means the controllers used there are analog controllers, in which both input and output are in analog form. But in many industrial process control applications, the signals is available in sequence of pulses or it is easier to use the signal in the form of sequence of pulses. With increase complexity of control systems, the digital controller is becoming more and more useful nowadays. They can use a 16-bit or 32bit word with a very high speed of 300 MHz, and can handle a large amount of data in any complex control process. The typical control system having digital controller (instead of analog controller) is called sampled data control system. Figure 11.1 shows the block diagram of such system.

Fig 11.1

r (t), e (t), g (t) and c (t) are analog signals or continuous signals. e* (t) and g*(t) are discrete signals. The overall system is hybrid in which the signal is in sampled form in digital controller and in continuous form in rest of the system. Such a system is sampled data control system. A digital control system makes use of digital computer, as shown in Fig. 11.2.

Fig 11.2

Sampled Data Control System 261

r* (t), e* (t) and g*(t) are in digital form or discrete form g (t) and c (t) are in continuous form. Discrete data control system is often used to represent both sampled data and digital control system.

11.2

COMPONENTS OF SAMPLED DATA CONTROL SYSTEM

The various components of sampled data control system as shown in Fig. 11.1, are explained below: (i) Sampler: It converts continuous time signal to a sequence of pulses. (ii) A to D converter (ADC): It transforms the analog signal to digital signal, which is expressed in some numerical code like binary code, hexadecimal code. (iii) Digital controller: All kinds of manipulations or the desired control action on the sampled input signal is taken up by digital controller. (iv) D to A converter (DAC): It decodes the numerical coded output of digital controller to continuous time signal. (v) Hold circuit: It holds the signal between two consecutive sampling instants by utilizing a proper circuit. (vi) Plant: It is the main system that is requires to be controlled by sampled data control system.

11.3

ADVANTAGES OF SAMPLED DATA CONTROL SYSTEM OVER ANALOG CONTROL SYSTEM

(i) Digital control system or sampled data control system enables time sharing between different input signals using same control equipment. (ii) Sampled data control system reduces the cost of equipment. (iii) In digital controllers or in digital computers, control function can be easily modified by changing a few program instructions. (iv) Power consumption of the sampled data control system is low as compared to that of analog control system. (v) Information is processed faster than that in analog control system.

11.4

APPLICATIONS OF SAMPLED DATA CONTROL SYSTEM

(i) In long distance data transmission. (ii) Driving of load at low power input signal. (iii) When output is essentially in discrete form like in radar tracking system.

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11.5

SAMPLER

It is the most basic element of sampled data control system. It samples the continuous signal to a sequence of pulses appearing at regular intervals of time. Sampling means that the signal at the output of the sampler is in the form of short duration pulses where each pulse is followed by a non-signal period in which no signal is observed. The advantage of doing sampling to the signal is that power demand made on the signal is reduced. So, it is beneficial for the signals which have weak power origin. So, in some cases, sampling process is purposely introduced. But in some cases, signals are available in sampled form like in radar. Figures 11.3 and 11.4 show the sampler and the sampling process respectively. f (t)

T

f *(t)

Sampler

Fig. 11.3

Fig. 11.4

f (t) is the continuous signal acting as input to the sampler. f *(t) is the sampled signal acting output of the sampler. T is the sampling time of sampler. The switch (sampler) is closed for very short duration Dt and then remains open for some duration. This process repeats after time period T, known as sampling time. Here, pulse width is Dt. Usually, sampling is made at regular intervals of time and that is called uniform periodic sampling. Now, if T is made very large, the sampling frequency 1/T would be too small and the information carried by input signal might be lost in the output signal. So, T must have a right value to avoid this.

Sampled Data Control System 263

11.5.1

Ideal Sampler

The pulse width of ideal sampler approaches to zero and the output f *(t) of ideal sampler becomes impulse train. The output of ideal sampler is a modulated waveform where the carrier is unit impulse train dT (t) and is being modulated by f (t) at sampling instants. This sampling process is shown in Figs. 11.5(a), (b), (c) and (d).

Fig. 11.5

The output of ideal sampler is given by f *(t) = f (t). dT (t)

(11.1)

Now, f *(t) is a uniform periodic sampled signal. Mathematically, the unit impulse train is •

dT (t) =

 d (t - kT )

(11.2)

k=0

where, d (t – kT ) is the unit impulse appearing at t = kT, k = 0, 1, 2, º Now, f (t) appears only at sampling instants, so f (t) can be represented as f (kT). Thus, sampled function, •

f *(t) = is a train of impulses.

 f ( kT ) d (t - kT )

k=0

11.5.2 Laplace Transform of f *(t) From equation (11.3),

•

f *(t) =

 f ( kT ) d (t - kT )

k=0

(11.3)

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Taking Laplace transform on both sides, È • ˘ F * (s) = L Í Â f ( kT ) d (t - kT )˙ ÍÎ k = 0 ˙˚ •

=

 f ( kT ) L ÎÈd (t - kT )˚˘

k=0

Èd

•

=

˘

 f ( kT ) L ÍÎ dt u (t - kT )˙˚

k=0 •

=

 f ( kT ) sL ÈÎu (t - kT )˘˚

k=0

1

•

=

 f ( kT ) s s e

k=0 •

= Thus,

F *(s) =

- skT

 f ( kT ) e

- skT

 f ( kT ) e

- skT

k=0 •

(11.4)

k=0

This F *(s) is called Laplace transform of starred function f *(t) and is known as impulse train Laplace transform. Suppose f (t) is sampled twice as shown in Fig. 11.6.

Fig. 11.6

If two samplers are synchronized that is their sampling times T and instant of sampling is same, then output of second sampler is again f * (t), i.e., [f * (t)]* = f * (t)

11.6 From equation (11.4)

•

F *(s) = Substituting

 f ( kT ) e

- skT

k=0

1 ln Z where Z > 1 T f ( kT ) Z - k

esT = Z or s = •

then,

Z-TRANSFORM

Z [f (t)] =

Â

k=0

(11.5)

Sampled Data Control System 265

The relation given in equation (11.5) is called Z-transform. •

Thus,

 f ( kT ) Z

Z [ f (t)] = F (Z) =

-k

(11.6)

k=0

Therefore, F *(s) can be converted into F (Z) by putting Z = e sT.

11.6.1 Z-transform of Some Useful Functions (i)

Unit step function: f (t) = u (t) = 1; t = kT •

Thus,

F (Z) = Z [1] = •

=

 f ( kT ) Z

-k

k=0

 (1) Z

, Z = esT

-k

k=0

= Z– 0 + Z– 1 + Z– 2 + º = 1 + Z– 1 + Z – 2 + º = So, (ii)

F (Z) = Z [1] =

Z Z-1

(11.7)

Unit ramp function: f (t) = t; t = kT •

Thus,

Z [t] =

 ( kT ) Z

k=0

-k

, Z = esT

= = (Z – 1) Z [t] = = =

0Z– 0 + TZ– 1 + 2 TZ– 2 + º. 0 + TZ–1 + 2 TZ–2 + º fi (Z – 1) T [Z– 1 + 2 Z– 2 + º] T (1 – Z – 1 + 2 Z – 1 – 2 Z– 2 + 3 Z– 2 + º) T (1 + Z– 1 + Z– 2 + º) 1 TZ = =T -1 Z-1 1-Z TZ So, F (Z) = Z [t] = (Z - 1)2 (iii) Exponential function: f (t) = e – at; t = kT •

Thus,

Z [e – at] =

 (e

k=0

- akT

) Z - k , Z = e sT

= 1 + e – aT Z–1 + e – 2 aT Z– 2 + º 1 = - aT 1 - e Z -1

(11.8)

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So,

F(Z) = Z[e-at] =

Z Z - e - aT

(11.9)

Table 11.1 shows the Laplace and Z-transform to some useful time functions. Table 11.1 Time function

Laplace transform

Z-transform

Unit step

1 s

Z Z–1

Unit ramp (t)

1 s2

TZ (Z – 1)2

1 s+a

Z Z – e– aT

Exponential (e – at )

1 s3

t2 2 tn n! te – at

1 – e – at

sin wt

e – at sin wt

cos wt

1 s

n+ 1

1 ( s + a )2 a s (s + a) s s2 + w 2 w ( s + a )2 + w 2 s s + w2 2

e – at cos wt

s ( s + a )2 + w 2

T 2 Z (Z + 1) 2 (Z – 1)3 n

lim

( – 1)

∂n È Z ˘ n ! ∂a n ÍÎ Z – e – a T ˙˚

aÆ0

TZ e – aT – aT 2

(Z – e ) (

Z 1 – e – aT

(

)

( Z - 1) Z – e– aT

)

Z sin wT Z 2 – 2 Z cos wT + 1 Z e – aT sin wT Z – 2 Z e – aT cos wT + e – 2 aT 2

Z ( Z – cos wT ) Z 2 – 2 Z cos wT + 1 Z 2 – Z e – aT cos wT Z – 2 Ze – aT cos wT + e –2 aT 2

Example 11.1 Find impulse train transform for the following: (a) f (t) = u (t) (b) f (t) = e – at Z (c) F (Z) = Z – e– aT

Sampled Data Control System 267

Solution: Step 1: (a)

f (t) = u (t) •

F *(s) =

 f ( kT ) e

k= 0

= 1 + e–sT + e–2 sT + º 1 = 1 – e– sT f (t) = e – at; t = kT

Step 2: (b)

•

F *(s) = = =

 f ( kT ) e

k= 0 •

Â

e– akT e – skT

Â

e– kT ( a + s )

k=0 •

Step 3:

To find F *(s), put Thus,

k=0

F *(s) =

esT 1 = – sT – aT sT -aT e –e 1– e e

Find Z-transform:

(b) f (t) = (1 – e– 5 t); T = 0.2 s (c) f (t) = sin wt (d) f (t) = e– at cos wt

f (t) = u (t) – e – 2 t Z [f (t)] = Z [u (t)] – Z [e– 2 t]

fi

•

Now,

(Answer)

Z = e+ sT

(a) f (t) = u (t) – e-2t

Solution: Step 1: (a)

(Answer)

– skT

= 1 + e –T (a + s) + e– 2 T(a + s) + º 1 F *(s) = – T (a + s) 1– e Z F (Z) = Z – e– aT

fi

Example 11.2

– skT

Z [u (t)] =

 (1) Z

–k

k=0

=1+Z–1+Z–2+º =

1 Z = 1 – Z –1 Z – 1

(Answer)

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and

Thus,

Z [e – 2 t] =

So,

So,

k=0

Z 2 – Z e–2 T – Z 2 + Z Z (1 – e– 2 T ) = – 2T (Z - 1)(Z - e ) (Z – 1) (Z – e– 2 T ) Z Z – Z – 1 Z – e– 5T

Z [f (t)] = –

0.632 Z Z – 1.368 Z + 0.368

(Answer)

2

f (t) = sin wt Z [f (t)] = Z [sin wt] = =

1 2j

•

Â

e j w kT Z – k -

k=0

1 2j

•

Âe

– j w kT

Z– k

k=0

˘ 1 È Z Z – Í j wT – jwT ˙ 2 j ÎZ – e Z–e ˚

– e– j w T + e j w T Z È = 2jÍ ÍÎ Z – e j w T Z – e– j w T

(

)(

)

˘ ˙ ˙˚

Ê e j w T – e– j w T ˆ 1 Z = Á ˜¯ Z 2 – Ze j w T – Z e – jw T + 1 2j Ë

(

= Z sin wT

Êe Z2 – 2 Z Á Ë Z sin wT = 2 Z – 2 Z cos wT + 1

Step 4:

(Answer)

f (t) = (1 – e – 5 t); T = 0.2 s

= Step 3:

e– 2 kT Z – k

= 1 + e –2 T Z – 1 + e – 4 T Z – 2 + º 1 Z = = – 2T –1 1– e Z Z – e– 2 T Z Z Z [f (t)] = – Z – 1 Z – e– 2 T =

Step 2: (b)

Â

1 – jwT

f (t) = e–at cos wt •

Z [ f (t)] =

Âe

k=0

– akT

)

cos ( k w T ) Z – k

+ e – jwT ˆ ˜¯ + 1 2

(Answer)

Sampled Data Control System 269

È • – akT jk w T – k • – akT – jkwT – k ˘ Z +Â e e Z ˙ ÍÂ e e ÍÎ k = 0 ˙˚ k=0

=

1 2

=

˘ 1È Z Z + Í j wT – aT – aT – j w T ˙ 2 ÎZ – e e Z–e e ˚

=

Z 2

È 2 Z – e– aT e j w T – e – aT e – j w T ˘ Í 2 j wT – j wT – aT –2 aT ˙ )+ e Î Z – Z e (e + e ˚

j wT È + e – jwT ˆ ˘ 1 – aT Ê e ˙ = Z ÍZ - e Á ˜ jwT 2 Ê e + e – jwT ˆ – 2 aT Ë ¯˚ 2 Î Z – 2 Z e– aT Á ˜¯ + e 2 Ë – aT Z (Z – e cos wT ) = 2 Z – 2 Z e– aT cos wT + e– 2 aT

Thus,

Z [e – at cos wt] =

Example 11.3

Z (Z – e– aT cos wT ) Z 2 - 2 Z e– aT cos wT + e– 2 aT 1 , where T = 1 s. s +2s+2

Find Z-transform of F (s) =

2

Solution: Step 1: Now, Thus,

Step 2: Thus, Step 3: Now, So,

fi

F (s) =

(Answer)

K1 K2 1 1 = = + s 2 + 2 s + 2 ÈÎs + ( 1 – j )˘˚ ÈÎs + ( 1 + j )˘˚ s + (1 – j ) s + (1 + j )

K1 = 1/2 j and K2 = – 1/2 j F (s) =

f (t) = F (Z) =

1 2 j ÈÎs + ( 1 – j )˘˚

–

1 2 j ÈÎs + ( 1 + j )˘˚

1 – (1 – j ) t 1 – (1 + j ) t e e – 2j 2j 1 Z 1 Z – 2 j Z – e– (1 – j ) T 2 j Z – e–(1+ j ) T

(

)

(

)

T = 1 s. F (Z) =

˘ Z È Z – e– (1 + j ) – Z + e– (1– j ) Í 2 (– 1+ j ) (– 1 – j ) –2 ˙ 2 j Î Z – Z (e +e )+ e ˚

(

)

˘ e– 1 e j + e– j Z È Í 2 ˙ F (Z) = 2 j Í Z – Z e–1 ( e j + e – j ) + e – 2 ˙ Î ˚

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Automatic Control Engineering

–1 = Ze

È ˘ Ê e j – e– j ˆ Í ˙ ÁË 2 j ˜¯ Í ˙ j –j Í ˙ Ê ˆ + e e –2 Í Z 2 – 2 Z e– 1 Á ˙ + e Ë 2 ˜¯ ÍÎ ˙˚

Z e– 1 sin 1 F (Z) = 2 Z – 2 Z e– 1 cos 1 + e – 2

fi Example 11.4

Find F (Z) whose F (s) =

(Answer)

2 e– 0.2 s 3 with T = 0.05 s. + s 2 + 9 s (s + 2)

Solution: Step 1:

2 e– 0.2 s 3 + s + 9 s (s + 2) 3 1 + 2 e– 0.2 s = 2 s + 32 s (s + 2)

F (s) =

2

= F1 (s) + F2 (s) . F3 (s) 3 s 2 + 32

Step 2:

F1 (s) =

fi

f1 (t) = sin 3 t Z sin 3 T F1 (z) = 2 Z – 2 Z cos 3 T + 1

So, Step 3: Now, Thus, Step 4: So, Thus,

Ê 1ˆ – 0.2 T Á ˜

ËT¯ – 0.2 s =2e F2 (s) = 2 e Z = e sT F2 (Z) = 2 Z– 0.2/T K K 1 1 /2 1 /–2 = 1+ 2 = + F3 (s) = s (s + 2) s s+2 s s+2

1 2 1 F3 (Z) = 2

f3 (t) =

Step 5: Therefore, F (Z) =

=

1 –2t e 2 Ê Z ˆ 1Ê Z ˆ ÁË ˜– Á ˜ Z – 1 ¯ 2 Ë Z – e– 2 T ¯ –

Z sin ( 0.15) 2 Z– 4 + Z 2 – 2 Z cos ( 0.15) + 1 2 Z – 1 0.149 Z

(

1 ˘ È 1 ÍÎ Z – 1 – Z – e– 0.1 ˙˚

1 È 1 ˘ + Z– 3 Í – Z – 1.97 Z + 1 Î Z – 1 Z – 0.905 ˙˚ 2

)

Sampled Data Control System 271

=

F (Z) =

fi

0.149 Z

(

È ˘ 0.095 + Z– 3 Í 2 ˙ Z – 1.97 Z + 1 Î Z – 1.905 Z + 0.905 ˚

)

2

0.149 Z 6 – 0.283 Z 5 + 0.135 Z 3 + 0.095 Z 2 – 0.188 Z + 0.095

(

Z 3 Z 4 – 3.89 Z 3 + 5.67 Z 2 – 3.69 Z + 0.905

) (Answer)

11.7

HOLD CIRCUIT/SIGNAL RECONSTRUCTION

After processing the sampled data signal by digital controller, it is necessary to convert it into an analog form for its utilization in the continuous part of the rest of the system. This operation is done by using different types of hold circuits, which are also called extrapolators. The use of hold circuit enables to hold the signal between two consecutive sampling instants at the preceded value till the next sampling instant is reached. The simplest hold circuit is called zero-order-hold (ZOH). This circuit holds output signal at a fixed level between two consecutive sampling instants such that the slope of the hold output signal is zero. In this, hold signal is zeroth derivative of an impulse. Figure 11.7 shows the sampler with ZOH.

Fig. 11.7

where, fh (t) is reconstructed output signal of ZOH. Figure 11.8(a), (b) and (c) show f (t), f *(t) and fh (t) respectively.

11.7.1 Unit Impulse to ZOH Figure 11.9 shows the ZOH, when unit impulse is given to it as input signal. Here, h (t) = u (t) – u (t – T) Taking Laplace transform on both sides, 1 1 - sT L [h (t)] = - e s s Also, L [d (t)] = 1 L [ h (t )] Thus, transfer function of ZOH i.e., Gho (s) = L [d (t )] 1 1 - sT - e s s fi Gho (s) = 1 1 1 - sT fi Gho (s) = - e s s

(11.10)

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Fig. 11.8

Fig. 11.9

11.8

PULSE TRANSFER FUNCTION

It is also called Z-transfer function. Consider a block diagram shown in Fig. 11.10.

Fig. 11.10

Here, fi

C (s) = G (s) R* (s) C *(s) = G *(s) R*(s)

Sampled Data Control System 273

C * (s) = G *(s) R * (s) C (Z ) G (Z) = R (Z )

fi fi

(11.11)

Thus, G (Z) is called pulse transfer function or Z-transfer function.

11.8.1 Rules to Reduce the Block Diagram Rule I: Blocks in cascade (a) When two blocks in cascade each having a sampler in its input, as shown in Fig. 11.11.

Fig. 11.11

Here,

C1(s) = R* (s) G1 (s)

and

C (s) = C1* (s) G2 (s)

fi

C (s) = R* (s) G1* (s) G2 (s)

fi

C *(s) = R* (s) G1*(s) G2* (s) C (Z ) = G1 (Z) G2 (Z) R (Z ) Equation (11.14) gives pulse transfer function as shown in Fig. 11.12.

Thus,

(11.12)

Fig. 11.12

(b) When two blocks in cascade having sampler in the input of the first block, as shown in Fig. 11.13.

Fig. 11.13

Here,

C (s) = R* (s) G1 (s) G2 (s)

fi

C *(s) = R* (s) [G1 (s) G2 (s)]*

fi

C (Z ) = [G1 G2](Z) R (Z )

(11.13)

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Equation (11.13) gives Pulse transfer function as shown in Fig. 11.14.

Fig. 11.14

(c) When two blocks in cascade having sampler in the input of second block, as shown in Fig. 11.15.

Fig. 11.15

Here, fi fi Rule 2: (a)

C (s) = [R G1 (s)]* G2 (s) C*(s) = [R G1 (s)]* G2* (s) C (Z) = [R G1](Z) G2 (Z)

(11.14)

Blocks in feedback loop

Error signal sampling

Fig. 11.16

Here, and fi fi fi

C (s) E (s) E (s) E*(s) E*(s)

= G (s) E*(s) = R (s) – C (s) H (s) = R(s) – G (s) E*(s) H (s) [from equation (11.15)] = R*(s) – E*(s) (GH)*(s) [1 + (GH)* (s)] = R*(s)

fi

E*(s) =

R * (s) 1 + (GH ) * (s )

Now, C *(s) = G *(s) E *(s) [from equation (11.15)] Putting E *(s) from equation (11.18), G *(s ) R * (s ) 1 + (GH ) * (s )

fi

C *(s) =

fi

C *(s ) G *(s ) = R *(s ) 1 + (GH )* (s )

(11.15)

(11.16)

Sampled Data Control System 275

fi

(b)

C (Z ) G (Z ) = R (Z ) 1 + (GH ) (Z ) Equation (11.17) gives the pulse transfer function.

(11.17)

Error and feedback sampling

Fig. 11.17

Here, C (s) = G (s) E*(s) fi C *(s) = G* (s) E*(s) and E (s) = R (s) – C *(s) H (s) Putting equation (11.18) in (11.19), fi E (s) = R(s) – G*(s) E*(s) H (s) fi E*(s) = R *(s) – G *(s) E*(s) H*(s) fi E*(s) [1 + G*(s) H*(s)] = R*(s) R*(s ) fi E*(s) = 1 + G *(s ) H * (s )

(11.18) (11.19)

(11.20)

Now, putting equation (11.20) in equation (11.18), G *(s ) R * (s ) C*(s) = 1 + G *(s ) H * (s ) C *(s ) G *(s ) fi = R *(s ) 1 + G *(s ) H *(s ) fi

C (Z ) G (Z ) = R (Z ) 1 + G (Z ) H ( z)

(11.21)

Equation (11.21) gives the pulse transfer function. (c) Input sampling

Fig 11.18

Here, and

C (s) = G (s) E (s) E (s) = R*(s) – H (s) C (s)

(11.22) (11.23)

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Putting equation (11.23) in equation (11.22), C (s) = G (s) R*(s) – G (s) H (s) C (s) fi C (s) [1 + G (s) H (s)] = G (s) R*(s) G (s) R*(s) fi C (s) = 1 + G (s) H (s) *

Ê G ˆ (s ) R*(s ) C*(s) = Á Ë 1 + GH ˜¯

fi

*

fi

C * (s) Ê G ˆ =Á ˜ (s) R * (s ) Ë 1 + GH ¯

fi

C (Z ) Ê G ˆ =Á ˜ (Z ) R (Z) Ë 1 + GH ¯

(11.24)

Equation (11.24) gives the pulse transfer function. Example 11.5 Find the pulse transfer function for the error sampled system shown in Fig. 11.19. R(s)

E(s) +

–

E*(s) T = 1s

1 s(s + 1)

C(s)

Fig. 11.19

Solution: Step 1: Now,

C (Z ) G (Z ) = R (Z ) 1 + (GH ) (Z )

G (s) =

1 1 1 = s (s + 1) s s + 1

Taking Laplace on both sides, G (t) = u (t) – e – t Taking Z-transform on both sides, G (Z) = Step 2: So,

0.632 Z Z Z Z Z = = –T Z–1 Z–e Z – 1 Z – 0.368 (Z – 1) (Z – 0.368)

(GH) (s) =

1 1 1 = s (s + 1) s s + 1

(GH) (Z) =

0.632 Z (Z – 1) (Z – 0.368)

Sampled Data Control System 277

0.632 Z (Z – 1) (Z – 0.368) 0.632 Z = 2 Step 3: Pulse transfer function = 0.632 Z Z – Z – 0.368 Z + 0.368 + 0.632 Z 1+ (Z – 1) (Z – 0.368)

fi Pulse transfer function = Example 11.6 function.

0.632 Z Z – 0.736 Z + 0.368

(Answer)

2

A closed loop control system is shown in Fig. 11.20, find pulse transfer

Fig. 11.20

Solution: Step 1: Now, So, Now, Thus,

C (Z ) G (Z ) = R (Z ) 1 + (GH ) (Z )

G ho =

1–

– sT

Ê 1 – e– sT ˆ G (s) = Á Ë s ˜¯

–s Ê 1 ˆ Ê1– e ˆ ÁË ˜¯ = Á s+1 Ë s ˜¯

–s Ê 1 ˆ 1– e ÁË ˜¯ = s+1 s (s + 1)

1 1 1 = s (s + 1) s s + 1 1 ˘ È1 G (s) = (1 – e – s) Í s s + 1 ˙˚ Î

Taking Z-transform on both sides, Z Z ˘ G (Z) = 1 – Z –1 ÈÍ – –T Î Z – 1 Z – e ˙˚

(

)

Z ˘ –1 È Z – = 1–Z Í – 1 Z Z e– T ˙˚ – Î

( (

= 1–Z Step 2:

)

–1

)

1 – e– s (GH) (s) = 2 s (s + 1)

È 1 – e–1 ZÍ ÍÎ ( Z – 1) Z – e– 1

(

( ) (

) )

˘ 1 – e– 1 ˙= ˙˚ Z – e – 1

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1 1 ˘ –s È 1 = 1– e Í 2 - + s s + 1 ˙˚ Îs

(

)

Taking Z-transform on both sides, Ê TZ Z Z ˆ (GH) (Z) = 1 – Z – 1 Á – + 2 –T ˜ Ë ( Z – 1) ( Z – 1) Z – e ¯

(

So,

)

–1 –1 Ê Z Z Z ˆ Ze – 2 e + 1 –1 = 1 – Z – + (GH) (Z) = Á 2 –T ˜ –1 Ë ( Z – 1) ( Z – 1) Z – e ¯ ( Z – 1) Z – e

(

)

(

(1 – e ) (Z – e )

)

–1

Step 3:

C (Z ) = R (Z )

–1

1+

11.9

( 1 – e ) ( Z – 1) –1

Ze– 1 – 2 e – 1 + 1

=

Z 2 – Z – e–1 + 1

(Answer)

( Z – 1) ( Z – e – 1 ) STABILITY ANALYSIS

Consider the block diagram shown in Fig. 11.21. R(s)

E(s) +

–

E*(s) T

G (s )

C(s)

H(s)

Fig. 11.21

Here,

C * (s) G *(s ) = R * (s ) 1 + (GH )* (s )

Characteristic equation becomes 1 + (GH)* (s) = 0. For stability, the roots of this characteristic equation must lie in the LHS of s-plane. In terms of Z-transform, the overall transfer function is C (Z ) G (Z ) = R (Z ) 1 + (GH ) (Z ) The characteristic equation is 1 + (GH) (Z) = 0. In terms of Z-transform, the stability region in Z-plane corresponding to s-plane is located by mapping from s-plane to Z-plane.

11.9.1

Mapping from s-plane to Z-plane

Z-transform variable ‘Z’ is defined as Z = e + sT

(11.25)

Sampled Data Control System 279

In s-domain analysis of stability, imaginary axis divides the stable and unstable regions as shown in Fig. 11.22 (a). The corresponding stable and unstable regions in Z-domain can be obtained by putting s = ± jw in equation (11.25), and then plotting the values of Z so obtained in another complex plane known as Z-plane. Thus, equation (11.25) becomes, Z = e ± jwT fi Z = (cos wT ± j sin wT) Here, magnitude of Z i.e., |Z| = 1 And phase angle of Z i.e., –Z = ± wT If the values of Z are plotted in Z-plane as variable w is varied along imaginary axis in s-plane, then a circle having unit radius and centre at origin is obtained in Z-plane. This is shown in Fig. 11.22 (b).

Fig. 11.22

Now, let s = (– a ± j w), a point having negative real part and lying in LHS of s-plane. Thus, Z = e (– a ± jw) T = e – a T (cos wT ± j sin wT) Here, |Z| = e – aT and –Z = ± wT As T is positive, so |Z|< 1 So, the point (– a ± jw) with negative real part located in s-plane lies inside the unit circle in Z-plane. Again, let s = + a ± jw, a point having positive real part and lying in RHS of s-plane. Thus, Z = e(+ a ± jw) T = e + a T (cos wT ± j sin wT) Here,

|Z| = e+ a T and –Z = ± wT

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As T is positive, so |Z| > 1 So, the point (+ a ± jw) with positive real part located in s-plane lies outside the unit circle in Z-plane. Therefore, it is concluded that for a sampled data control system to be stable, all roots of Z-transformed characteristic equation must be located inside the unit circle centered at origin in z-plane. In case, any root lies outside the circle, the system becomes unstable. This is the general method of finding out stability of a control system, when it is represented in Z-domain. But this method is suitable for only lower ordered characteristic equations. If the characteristic equation of the system is of higher order (higher than two), then it is simple to determine the roots and hence stability. So, some methods are formalized to determine the stability.

11.9.2

Modified Routh’s Hurwitz Test

It has been studied that the control system is stable with respect to z-domain, if the roots of characteristic equation lie within the unit circle. The standard methods for finding stability, that were used in s-domain analysis like Routh’s criterion, Nyquist criterion can also be applied in z-domain if a complex transformation is made that transfers the interior of the unit circle of z-plane to the left hand side of a new plane, termed as say x-plane. The exterior of the unit circle of z-plane is mapped to the right hand side of s-plane. This transformation is called bilinear transformation. So, according to this transformation, z–1 x+1 or z = x= z +1 x–1 with z = e j q, q is varying from – p to + p through 0° in anticlockwise direction. Thus,

e j q – 1 cos q + j sin q – 1 = e j q + 1 cos q + j sin q + 1 q q cos q – 1 + j 2 sin cos 2 2 = q q cos q + 1 + j 2 sin cos 2 2 q q q – 2 sin 2 + j 2 sin cos 2 2 2 = q q 2 q 2 cos + j 2 sin cos 2 2 2

x=

q qˆ Ê – sin + j cos 2 2 ˜ tan q = ÁÁ q q ˜ 2 ÁË cos + j sin ˜¯ 2 2 q = j tan 2

Sampled Data Control System 281

For For For

Ê pˆ Ê pˆ q = – p, x = + j tan Á – ˜ = – j tan Á ˜ = – j • Ë 2¯ Ë 2¯ q = 0, x = 0 Ê pˆ q = p, x = j tan Á ˜ = + j • Ë 2¯

So, when q varies from – p to + p through 0°, x varies from – j • to + j • through 0°. So, stability region in z-plane can be mapped into x-plane as shown in Fig. 11.24 (a) and (b). So, stability conditions in x-plane are same as that in s-plane. Hence, same methods which were used in s-domain stability analysis can be used in z-domain stability analysis, after doing transformation. Thus, characteristic equation in z-plane can be z–1 , to get new characteristic equation in transformed into x-plane by putting x = z+1 x-plane. Then, stability can be found out using Routh’s Hurwitz criterion. Example 11.7 Find stability of the following sampled data control system for T = 1 s. R(s)

E(s) +

–

E*(s)

5 s(s + 0.5)

T = 1s

C(s)

Fig. 11.23

Solution: C (Z ) G (Z ) = Step 1: Transfer function in z-domain is R (Z ) 1 + (GH ) (Z ) 5 10 10 = Here, G (s) = s (s + 0.5) s s + 0.5 Taking Z-transform on both sides,

( (

)

10 Z 1 – e– 0.5 10 Z (1 – e– 0.5 ) Z È Z ˘ = = G (Z) = 10 Í – 0.5 T ˙ Z 2 – Z (1 + e– 0.5 ) + e– 0.5 ÎZ – 1 Z – e ˚ ( Z – 1) Z – e– 0.5 T

Step 2:

)

10 Z (1 – e– 0.5 ) (GH) (Z) = 2 Z – Z (1 + e– 0.5 ) + e– 0.5

10 Z (1 – e– 0.5 ) Step 3: Pulse transfer function = 2 Z – Z (– 9 + 11 e– 0.5 ) + e– 0.5

Step 4: Characteristic equation is Z2 – Z (–9 + 11 e– 0.5) + e – 0.5 = 0 fi

Z2 – 2.33 Z + 0.606 = 0

Step 5: Apply bilinear transformation, put z =

x+1 x–1

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Automatic Control Engineering 2

Ê x + 1ˆ Ê x + 1ˆ – 2.33 Á + 0.606 = 0 Characteristic equation becomes Á ˜ Ë x – 1¯ Ë x – 1 ˜¯

– 0.724 x2 + 0.788 x + 3.93 = 0

fi

Step 6: Applying Routh’s-Hurwitz criterion, Routh’s array becomes x 2 – 0.724 3.93 x 1 0.788 0 0 x 3.93

Step 7: As first term in the first column of Routh’s array is negative, system is unstable. Now, when the system given in Fig. 11.25 was linear continuous, the characteristic equation was, 1 + G (s) H (s) = 0 fi

5 = 0 fi s2 + 0.5 s + 5 = 0 s (s + 0.5) Routh’s array becomes

1+

s2 1 5 1 s 0.5 0 s0 3

As all the terms in first column are positive, the system is stable. Now, again suppose that the transfer function is taken as 1 G (s) = s (s + 1) (s + 2) È1 1 1 ˘ So, G (Z) = Z-transform of Í – + ˙ Î 2 s s + 1 2 (s + 2) ˚ È Z ˘ Z (0.594 + 0.4 Z ) Z Z = – + = Í –1 –2 ˙ 2 (Z – e ) ˚ 2 ( Z – 1) ( Z – 0.368) ( Z – 0.135) Î 2 ( Z – 1) Z – e

So, characteristic equation becomes, 1 + G (Z) = 0 fi Z 3 – 1.3 Z 2 + 0.85 Z – 0.5 = 0 x+1 Apply bilinear transformation, put z = x–1 3

2

x + 1ˆ Ê x + 1ˆ Ê x + 1ˆ Characteristic equation becomes ÊÁ – 1.3 Á + 0.85 Á – 0.5 = 0 Ë x – 1 ˜¯ Ë x – 1 ˜¯ Ë x – 1 ˜¯

fi

3.65 x3 + 1.95 x2 + 2.35 x + 0.05 = 0

(11.26)

Sampled Data Control System 283

Step 8: Applying Routh’s-Hurwitz criterion, Routh’s array becomes x3 x2 x1 x0

3.65 2.35 1.95 0.05 2.26 0 0.05

As all the terms in the first column of Routh’s array are positive, system is stable. Now, when the system given in equation (11.26) was linear continuous, the characteristic equation was, 1 + G (s) H (s) = 0 1 = 0 fi s3 + 3 s 2 + 2 s + 1 = 0 fi 1+ s (s + 1) (s + 2) Routh’s array becomes s3 s2 s1 s0

1 3 5/3 1

2 1 0

As all the terms in first column are positive, the system is stable. Hence, it is seen that if sampling and ZOH are applied to stable linear continuous systems, the resulting sampled system may become unstable. Thus, sampling has destabilizing effect on the control system.

11.9.3 Jury’s Stability Test Consider nth order characteristic polynomial, (11.27) F (z) = an zn + an – 1 zn – 1 + an – 2 zn – 2 + º + a0 = 0; an > 0 The first test for Jury’s stability test is the test for necessity. If the necessity test is satisfied, only then the second test that is sufficient condition test is needed to be performed. (i) Necessity test conditions for stability are F(1) >0 (– 1)n F(– 1) > 0 |a0| < an If above conditions are satisfied, then sufficient conditions are tested. (ii) Sufficient conditions are tested by two methods: (a) First method: It requires construction of Jury’s array. In this array, there are (2 n – 3) rows and first two rows are coefficients of characteristic polynomial. Elements of subsequent rows are defined by elements of first two rows. The array is constructed like

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Automatic Control Engineering

Row 1

a0

a1

a2

–

–

an – 1

an

2

an

an – 1

an – 2

–

–

a1

a0

3

b0

b1

b2

–

–

bn – 1

4

bn – 1

bn – 2

–

–

–

b0

5

c0

c1

c2

–

–

cn – 2

6

cn – 2

cn – 3

–

–

–

c0

– 2n – 3

– –

– –

– –

– –

– –

– –

where, bk =

a0 an

ck =

b0 bn – 1

an – k ; k = 0, 1, 2, 3, º, (n – 1) ak bn – k – 1 ; k = 0, 1, 2, 3, º, (n – 2) bk

The sufficient condition for stability is |a0| < |an|; |b0| > |bn – 1|; |c0| > |cn – 2|; º (b) Second method: Two triangularised matrices M and N are constructed using coefficients of characteristic equation given in equation (11.27). È Í Í M = ÍÍ Í ÍÎ

an 0 . . 0

È an – 2 Ía Í n–3 N= Í . Í Í . ÍÎ a0

an – 1 an

. an – 1

. .

. .

0

0

.

.

an – 3 an – 4

. .

. .

a1 a0

.

.

.

.

a2 ˘ a3 ˙˙ ˙ ˙ ˙ an ˙˚ ( n – 1) a2 ˘ 0 ˙˙ ˙ ˙ ˙ 0 ˙˚ ( n – 1)

M and N are square matrices of order (n – 1). Now, let P1 = M + N and P2 = M – N The system is asymptotically stable, if P1 and P2 are positive innerwise, means when all the determinants starting from the centre element and proceeding outwards to the total matrix must be positive.

Sampled Data Control System 285

Example 11.8 Consider the characteristic equation as 2 z4 + 8 z3 + 12 z2 + 5 z + 1 = 0. Determine stability using Jury’s test. Solution: Step 1: Firstly check the necessary condition: F (1) = 2 + 8 + 12 + 5 + 1 = 28 > 0 n (– 1) F (–1) = (–1)4 [2 (1) + 8 (–1) + 12 (1) + 5 (–1) + 1] = 2 > 0 and |a0| = 1 and |an|= 2. Thus, |a0| < |an| So, necessary conditions are satisfied. Step 2: Check the sufficient condition using first method. Make Jury’s array: a0 a4 – k ; k = 0, 1, 2, 3 bk = a4 ak Thus,

Again,

b0 =

1 2 =–3 2 1

b1 =

1 8 = – 11 2 5

b2 =

1 12 = – 12 2 12

b3 =

1 5 =–2 2 8

ck =

b0 b3

c0 =

–3 –2 =5 –2 –3

c1 =

– 3 – 12 =9 – 2 – 11

c2 =

– 3 – 11 = 14 – 2 – 12

b3 – k ; k = 0, 1, 2 bk

Row 1

1

5

12

8

2

2

2

8

12

5

1

3

–3

– 11

– 12

–2

4

–2

– 12

– 11

–3

5

5

9

14

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Step 3: Here, |a0| < |an| fi 1 < 2 (satisfied) |b0| > |bn – 1| fi 3 > 2 (satisfied) |c0| > |cn – 2|fi 5 < 14 (not satisfied) Hence, system is unstable. Step 4: Check the sufficient condition using second method: Triangularised matrices are developed as

and

Step 5: Now, Here,

and

È a4 Í M = Í0 ÍÎ 0

a3 a4 0

a2 ˘ È 2 8 12 ˘ ˙ Í ˙ a3 ˙ = Í 0 2 8 ˙ a4 ˙˚ ÍÎ 0 0 2 ˙˚

È a2 Í N = Í a1 ÍÎ a0

a1 a0 0

a0 ˘ È12 5 1˘ ˙ Í ˙ 0 ˙ = Í 5 1 0˙ 0 ˙˚ ÍÎ 1 0 0 ˙˚

(Answer)

È14 13 13˘ P1 = M + N = Í 5 3 8 ˙ Í ˙ ÍÎ 1 0 2 ˙˚ 5>0

È14 13 13˘ Í ˙ Í 5 3 8 ˙ = 14 ((6 – 0) –13 (10 – 8) + 13 (0 – 3) = 19 > 0 ÍÎ 1 0 2 ˙˚

Thus, P1 is positive innerwise. Step 6: Again, Here,

and

È– 10 3 11˘ Í ˙ P2 = M – N = Í – 5 1 8 ˙ ÍÎ – 1 0 2 ˙˚ 1>0

È– 10 3 11˘ Í ˙ Í – 5 1 8 ˙ = – 10 (2 – 0) – 3 (– 10 + 8) + 11 (0 + 1) = – 3 < 0 ÍÎ – 1 0 2 ˙˚

Thus, P2 is not positive innerwise. Hence, system is unstable.

(Answer).

Sampled Data Control System 287

Exercise 1. Discuss the concept of sampled data control system. Explain its various components. 2. Discuss advantages of sampled data control system over analog control system. State some applications of sampled data control systems. 3. Explain the function of sampler in sampled data control system in detail. Discuss ideal sampler. 4. Find Z-transform of unit step function, unit ramp function and exponential function. 5. Explain the process of signal reconstruction with the help of hold circuit. 6. Explain sample and hold. Obtain frequency response of ZOH. 7. Find Z-transform of (a) cos wt (b) e – at sin wt 8. Find pulse transfer function of the system shown in Fig. 11.24. R(s)

E(s) +

–

E*(s)

5/(s + 1)

T = 1s

C(s)

1/ss 1/

Fig. 11.24

9. Obtain the unit-step response of the system shown in Fig. 11.25. R( s )

E(s) +

–

E*(s) T = 1s

ZOH

1/s(s + 1)

C(s)

Fig. 11.25

10. Define stability of discrete control system. Explain Jury’s test of stability. 11. Explain with suitable example, how sampling has destabilizing effect on sampled data control system. 12. Consider the following characteristic polynomial. Apply Jury’s stability test to find stability of control system. F (Z) = 4 z4 + 6 z3 + 12 z2 + 5 z + 1 13. Explain Routh Hurwitz test for stability of discrete control systems using bilinear transformation.

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Automatic Control Engineering

Objective Type Questions 1. The overall pulse transfer function of the system shown is

(a)

1 – e– 1 Z – e– 1

(b)

Z (1 + e– 1 ) (Z – 1) (Z + e– 1 )

(c)

1 + e– 1 Z + e– 1

(d)

Z (1 + e– 1 ) (Z – 1) (Z – e– 1 )

2. For the given sampled-data system

(a) G1 G2 (Z)

(b) G1 (s) G2 (s)

(c) G2 (Z) G1 (Z)

(d) G2 (Z)

3. Which of the following stability test is used for sampled data control system? (a) Nyquist test (b) Bode plot test (c) Polar plot test (d) Jury’s test 4. Which of the following components of discrete control system converts continuous time signal to a sequence of pulses? (a) ZOH (b) DAC (c) Sampler (d) ADC 5. [f *(t)]* is equal to (a) f *(t) (b) f (t) (c) f **(t) (d) f (t)** 6. Z-transform of unit ramp function is Z TZ (a) Z – 1 (b) (Z – 1)2

(c)

T (Z – 1)2

(d)

T (Z – 1)

7. Transfer function of ZOH is

1 – e– sT (a) s 1 – e– sT (c) s2

(b) 1 – e – sT – sT (d) s – e s

Sampled Data Control System 289

8. Signal reconstruction is digital control system is done by using (a) DAC (b) ADC (c) ZOH (d) Sampler 9. Decoding of numerical coded output of digital controller to continuous time signal is done by (a) ADC (b) DAC (c) Sampler (d) ZOH

Answers 1. (a) 6. (b)

2. (c) 7. (a)

3. (d) 8. (c)

4. (c) 9. (b)

5. (a)

MATLAB Programs 1 to discrete transfer s + 2s + 2 function with sampling time 1s. Also, find back the continuous transfer function.

P1. Convert the continuous transfer function T (s) =

2

Program: nc = [0 1]; dc=[1 2 2]; Tc=tf(nc,dc) Ts=1; Td=c2d(Tc,Ts) T=d2c(Td)

% Numerator of continuous transfer function % Denominator of continuous transfer function % Continuous transfer function % Sampling time % Continuous to discrete conversion with Ts, default method is zero order hold % Discrete to continuous conversion

Execution: Transfer function: 1 ------------s^2 + 2 s + 2 Transfer function: 0.2458 z + 0.1231 ----------------------z^2 - 0.3975 z + 0.1353 Sampling time: 1 Transfer function:

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Automatic Control Engineering

1 ------------s^2 + 2 s + 2 z–1 P2 Find Laplace transform of the following z-transfer function F (z) = 2 z + z + 0.3 with sampling time 0.1 s. Also, find back the z- transfer function. Program: nd = [1 -1]; dd=[1 1 0.3]; Ts=1; Td=tf(nd,dd,Ts Tc=d2c(Td) Td=c2d(Tc,Ts)

% Numerator of z-transfer function % Denominator of z-transfer function % Sampling time % z-transfer function % Discrete to continuous conversion % Continuous to discrete conversion with Ts, default method is zero order hold

Execution: Transfer function: z-1 ------------z^2 + z + 0.3 Sampling time: 0.1 Transfer function: 121.7 s - 3.215e-012 --------------------s^2 + 12.04 s + 776.7 Transfer function: z-1 ------------z^2 + z + 0.3 Sampling time: 0.1 z–1 with sampling time 0.1 s. z 2 + z + 0.3 Resample it at 0.05 s. Also, sample it back at 0.1 s.

P3 Consider z-transfer function as F (z) =

Program: nd = [1 -1]; dd=[1 1 0.3]; Ts=1; Td1=tf(nd,dd,Ts) Td2=d2d(Td1,0.05) Td1=d2d(Td2,0.1)

% % % % % %

Numerator of z-transfer function Denominator of z-transfer function Sampling time z-transfer function with Ts = 1s z-transfer function with Ts = 0.05s z-transfer function with Ts = 0.1s

Sampled Data Control System 291

Execution: Transfer function: z-1 ------------z^2 + z + 0.3 Sampling time: 0.1 Transfer function: 3.237z - 3.237 ----------------------z^2 - 0.3089 z + 0.5477 Sampling time: 0.05 Transfer function: z-1 ------------z^2 + z + 0.3 Sampling time: 0.1 P4 Consider z-transfer function as F (z) =

Program: nd = [-0.4 -0.1] dd=[1 1.05 0.08]; Td=tf(nd,dd,-1) rlocus(Td)

% % % %

– 0.4 z – 0.1 . Draw root locus. z + 1.05 z + 0.08 2

Numerator of z-transfer function Denominator of z-transfer function z-transfer function with unspecified sampling time Root locus of given z-transfer function

Execution: Transfer function: -0.4 z - 0.1 ------------------z^2 + 1.05 z + 0.08 Sampling time: unspecified

Root Locus

1 0.8

Imaginary Axis

0.6 0.4 0.2 0 –0.2 –0.4 –0.6 –0.8 –1 –1

–0.5

0

0.5 Real Axis

1

1.5

2

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Automatic Control Engineering

– 0.4 z – 0.1 P5 Consider z-transfer function as F (z) = 2 . Draw root locus. Also, z + 1.05 z + 0.08 plot the z-plane grid on the root locus.

Program: nd = [-0.4 -0.1]; dd=[1 1.05 0.08]; Td=tf(nd,dd,-1) rlocus(Td) zgrid

% % % % %

Numerator of z-transfer function Denominator of z-transfer function z-transfer function with unspecified sampling time Root locus of given z-transfer function z-plane grid lines

Execution: Transfer function: – 0.4 z - 0.1 ------------------z^2 + 1.05 z + 0.08 Sampling time: unspecified

I

Laplace Transforms of Some Useful Functions

APPENDIX

Time function

Laplace transform

Unit step u (t)

1 s

Unit ramp (t)

1 s2

Unit Impulse (d (t)) Exponential (e ) – at

t2 2 tn

1

1 s+a 1 s3 n! sn + 1

te – at

1 ( s + a )2

tne – at

n! (s + a)n + 1

1 – e– at

a s (s + a)

1 (1 – e– at ) a

1 s (s + a)

1 (e– at – e– bt ) b–a

1 (s + a) (s + b )

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Automatic Control Engineering

1 (be – bt – ae– at ) b–a

s (s + a) (s + b )

(1 – at) e – at

s ( s + a )2

sin wt

w s + w2 2

e – at sin wt

cos wt

w ( s + a )2 + w 2 s s + w2 2

e – at cos wt

s ( s + a )2 + w 2

1 – cos wt

w2 s (s 2 + w 2 )

Answers to Unsolved Exercise (Selected Numericals) Chapter 2 6. 7.

8. 10.

( G1 G 2 + G6 ) ( G 3 G 4 G 5 ) 1 + G 4 G5 H 1 + G3 G 4 H 2 + G2 G3 G 4 H 2 G1 G 2 G 3 G 4 G6 - G1 G 2 G 3 + G 1 G 2 G 5 G 6 1 + G 2 G 3 H 1 + G1 G 2 H 2 G1 G2 G3 G4 G5 + G6 (1 + G2 H 1 + G4 H 2 + G2 G4 H 1 H 2 ) 1 + G 2 H 1 + G 4 H 2 + G1 G 2 G 3 G 4 G 5 H 3 - G6 H 3 + G 2 G 4 H 1 H 2 + G 2 G6 H 1 H 3 + G 4 G6 H 2 H 3

G1 G2 G3 + G4 (1 + G2 G3 H 2 + G2 H 1 - G1 G2 H 1 ) 1 + G 2 G 3 H 2 + G 2 H 1 - G1 G 2 H 1

Chapter 4 1.

0.495, 0.97

2. 0.192 3. 4.

5 2

s (6 + s ) + 10 ( s + 2)

, 0.5

2

2

,

2

( s + 2) + KK b ( s + 2) + KK b

Chapter 5 1. 1, 5 2. 0.2, 5 3. 20 5.

- KK b

( sK + b ) 2

Ès + s ( a - K ) ˘ Î ˚

6. 4.03, 1.72 9. 0.134, 190.537 10. 1, •, 14, 0, 0.357 11. 1/16, 4

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Automatic Control Engineering

Chapter 6 1.

3. 4. 5. 6.

(a) unstable (2) (b) unstable (2) (c) stable (d) stable 0 < K < 48.75 666.25, 4.06 rad/s stable 70 > K > 0

Chapter 7 2. 5. 6. 7.

– 4.2, – 71° and + 71°, K < 229.44 8.28 0.3 2.09

Chapter 8 3. wgc = 2 rad/s, wpc = 18 rad/s, GM = 24 db, PM = 31° 6. 119.7, 37.22 8. stable

References 1. RC Dorf and RH Bishop, Modern Control Systems, Addison Wesley Publishing Company. 2. BC Kuo, Automatic Control Systems, Prentice Hall, India. 3. Francis H Raven, Automatic Control Engineering, TMH Publishing Company Limited. 4. K Ogata, Modern Control Engineering, Prentice Hall, India. 5. SP Eugene Xavier, Principles of Control Systems, S Chand and Company Limited. 6. TS Chang, Chen CT, On the Routh’s-Hurwitz Criterion, IEEE Transactions on Auto Control, 1974, pp 250-251. 7. YH Ku, Analysis and Control of Linear Systems, International Text Book Co., Scranton, Pennsylvania. 8. David K. Cheng, Analysis of Linear Systems, Addison Wesley Publishing Company. 9. IJ Nagrath, M Gopal, Control Systems Engineering, Wiley Eastern Limited, New Delhi. 10. S. Hasan Saeed, Automatic Control Systems (with MATLAB Programs), SK Kataria & Sons, New Delhi. 11. RS Chauhan, Linear Control Systems, Umesh Publications, New Delhi.

Index

A Absolute stability 127 AC servomotors 231 AC tachogenerators 242 All pass systems 223 Analog control system 261 Armature controlled DC servomotors 233 B Bode plot 183 C Closed contour 208 Closed loop control system 2, 3, 5, 7 Control systems classification of 4 effect of feedback in 65 effect of feedback on disturbances in 73 effect of feedback on dynamics of 72 effect of feedback on stability of 73 gain of 71 modeling of 48 parameter variations 65 sensitivity of 66 D Damper 49 DC tachogenerators 241 Derivative controller 253 E Electrical systems modeling of 51 Error constants 109 Error detectors 237 Error response 81 F Field controlled DC servomotor 235

Field controlled motor limitations of 237 Frequency response 179 Frequency response analysis 179 Frequency response specifications 182 Friction 49 G Gain margin 192, 204, 216 H Hold circuit 271 Hurwitz stability criterion 128 Hydraulic systems 60 effects of feedback on 66 I Impulse signal 83 Integral controller 252 Inverse polar plot 207 J Jury’s stability test 283 M Mason’s gain formula 33 Mechanical systems modeling of 48 Minimum phase systems 223 Modified Routh’s Hurwitz test 280 N Non-minimum phase systems 223 Nyquist criterion 212 Nyquist stability criterion 208 O Open loop control system 1, 2 P Parabolic signal 83

300

Index

Performance indices 102 Permanent magnet stepper motor 243 Phase margin 192, 204, 216 PID controllers 251 tuning of 255 Polar plot 200 Poles cancellation of 168 Potentiometers 237 Proportional controller 252 Proportional-derivative controller 254 Proportional-integral controller 254 Proportional-integral-derivative controller 255 Pulse transfer function 272 R Ramp signal 82 Regenerative feedback 77 Relative stability 127 Relative stability analysis 141 Robust control systems 249 Robustness analysis of 251 Root contours 172 Root locus technique 146 Root locus construction rules of 147 Root sensitivity 250 Rotational mechanical system 50 Routh’s Hurwitz criterion special cases of 131 applications of 137 Routh’s stability criterion 129 S Sampled data control system 261 Sampler 262 Servomotors 231 Signal flow graph 30 Signal reconstruction 271 Stability 123, 204, 216 types of 127 Stability analysis 278

Steady state error 82 Steady state errors 103, 109 Steady state response 81 Step signal 82 Stepper motors 243 Synchro transmitter 239 Synchro transmitter-control transformer pair 240 Synchros 239 System sensitivity 249 T Tachogenerators 241 Thermal systems 58 Time response 179 Time response analysis 81 Time response of first order system 84 of second order system 89 specifications of second order system 96 to the unit parabolic input 86 to the unit ramp input 85 to the unit step input 84 to the unit ramp input 92 to the unit step input 89 Transfer function of a closed loop system 17 of electric circuits 14 representation of 12 terms related to 13 Transient analysis input signals for 82 Transient response 81 Translational mechanical systems 48 V Variable reluctance stepper motor 244 Z Zeros cancellation of 168 Ziegler-Nichols method 256 Z-transform 264

Automatic Control Engineering

Aditi Gupta is Assistant Professor in the Department of Electrical & Electronics Engineering at UIET, Panjab University, Chandigarh. She received her B.Tech. degree in Electrical Engineering from MMEC, Mullana, Kurukshetra University, Kurukshetra in 2002 with honours and Masters degree in Electrical Engineering from Panjab University, Chandigarh in 2004 with distinction. She is pursuing Ph.D in Electrical Engineering from Panjab University, Chandigarh. Her research interest includes automatic generation control of power systems, wind power, power system deregulation and congestion management. She has contributed several research papers in national and international journals.

978-93-89583-74-8

1

2

E

c(t) ess r(t)=1

Aditi Gupta • Yajvender Pal Verma

Yajvender Pal Verma received his B.Tech. degree in Electrical Engineering from NIT Hamirpur in 2000 with Honours and Masters degree in Power Systems from Panjab University, Chandigarh in 2002. He is Ph.D in Electrical Engineering from NIT Kurukshetra, Haryana, India. He is presently working as Assistant Professor in the Department of Electrical & Electronics Engineering at UIET, Panjab University, Chandigarh. His research interest includes distributed generation, control and operation of wind power, power system restructuring, congestion management and power system optimisation. He has authored more than 40 research papers in national and international journals of repute and conferences. He has also carried out two research projects sponsored by Department of Science & Technology. He is a member of IEEE, CSI and IAENG.

Automatic Control Engineering

AUTOMATIC CONTROL ENGINEERING

The book presents the topics of control engineering in a relatively easy language, which is of great importance for the beginners. It is compulsory for every student to start with the basic concepts and gradually reach to the competitive level. This book includes competitive objective-type questions at the end of each chapter. Moreover, the major complex topics of control engineering, such as Root Locus Technique, Bode Plot and Nyquist Criterion are explained in an easy manner using step-by-step method. In addition to this, some advanced topics of control engineering, such as sampled data control systems and robust control systems are also included in this book.

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